Mixing
how to pass multiple variables from main function down to other functions C++
I am stuck on this program as follows:
Write a program that takes as input five numbers and outputs the mean (average) and standard deviation of the numbers. If the numbers are x1,x2,x3,x4,x5.
To solve mean : mean = (x1 + x2 + x3 + x4 + x5) / 5;
To solve deviation:
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
Your program must contain at least the following functions: a function that calculates and returns the mean and a function that calculates the standard deviation.
I am trying to use variable x1,x2,x3,x4,x5 from main but I am getting unitialized local variable errors and can not compile it.
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
return 0;
} // end function main
double meanfn()
{
double x1, x2, x3, x4, x5;
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
double x1, x2, x3, x4, x5;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
c++ function
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
add a comment |
I am stuck on this program as follows:
Write a program that takes as input five numbers and outputs the mean (average) and standard deviation of the numbers. If the numbers are x1,x2,x3,x4,x5.
To solve mean : mean = (x1 + x2 + x3 + x4 + x5) / 5;
To solve deviation:
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
Your program must contain at least the following functions: a function that calculates and returns the mean and a function that calculates the standard deviation.
I am trying to use variable x1,x2,x3,x4,x5 from main but I am getting unitialized local variable errors and can not compile it.
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
return 0;
} // end function main
double meanfn()
{
double x1, x2, x3, x4, x5;
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
double x1, x2, x3, x4, x5;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
c++ function
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
3
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54
add a comment |
I am stuck on this program as follows:
Write a program that takes as input five numbers and outputs the mean (average) and standard deviation of the numbers. If the numbers are x1,x2,x3,x4,x5.
To solve mean : mean = (x1 + x2 + x3 + x4 + x5) / 5;
To solve deviation:
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
Your program must contain at least the following functions: a function that calculates and returns the mean and a function that calculates the standard deviation.
I am trying to use variable x1,x2,x3,x4,x5 from main but I am getting unitialized local variable errors and can not compile it.
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
return 0;
} // end function main
double meanfn()
{
double x1, x2, x3, x4, x5;
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
double x1, x2, x3, x4, x5;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
c++ function
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
I am stuck on this program as follows:
Write a program that takes as input five numbers and outputs the mean (average) and standard deviation of the numbers. If the numbers are x1,x2,x3,x4,x5.
To solve mean : mean = (x1 + x2 + x3 + x4 + x5) / 5;
To solve deviation:
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
Your program must contain at least the following functions: a function that calculates and returns the mean and a function that calculates the standard deviation.
I am trying to use variable x1,x2,x3,x4,x5 from main but I am getting unitialized local variable errors and can not compile it.
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
return 0;
} // end function main
double meanfn()
{
double x1, x2, x3, x4, x5;
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
double x1, x2, x3, x4, x5;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
c++ function
c++ function
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
asked Nov 20 '18 at 3:50
Wessley Ray ConeWessley Ray Cone
12
12
3
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54
add a comment |
3
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54
3
3
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54
add a comment |
2 Answers
2
active
oldest
votes
You have two problems of understanding basics in C++:
- Variable scope.
- Function parameters.
To briefly explain:
Variables in C++ live only in the scope they were declared in. In your case, you declared: x1, x2, x3, x4, x5 inside of the function main which means they do not exist outside that scope. The variables x1, x2, x3, x4, x5 in meanfn and deviationfn are completely different variables. They only share the same names and that's about it. Therefore, when you declared them inside the functions, they had no previously assigned values which means you cannot use them.
As for function parameters, you have to declare what parameters a function accepts in its signature. Your meanfn signature is double meanfn() which accepts no parameters. If you want it to accept 5 and only 5 double variables, you should change it to accommodate.
There are two solutions based on the explanations above:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
// Variable declarations:
double x1, x2, x3, x4, x5; // outside of 'main' making them global and everything has access to these variables.
int main()
{
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// CALL YOUR FUNCTIONS, THEY WONT BE CALLED BY THEIR OWN
double mean = meanfn();
double deviation = deviationfn(mean);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn()
{
double mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
In the solution above, the variables x1, x2, x3, x4, x5 were moved to be global variables making their access global and hence all functions can use them.
The second solution would consider parameter passing for the functions, such as:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn(double x1, double x2, double x3, double x4, double x5);
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5);
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// see the difference in the function calls?
double mean = meanfn(x1, x2, x3, x4, x5);
double deviation = deviationfn(mean, x1, x2, x3, x4, x5);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn(double x1, double x2, double x3, double x4, double x5)
{
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
add a comment |
Rather than having 5 variables with numbered names, you can instead have one variable that contains all 5.
#include <iostream>
#include <cmath>
#include <array>
using Values = std::array<double, 5>;
double mean(Values xs)
{
return (xs[0] + xs[1] + xs[2] + xs[3] + xs[4]) / xs.size();
}
double distribution(Values xs, double x_bar)
{
return sqrt((pow(xs[0] - x_bar, 2) + pow(xs[1] - x_bar, 2) + pow(xs[2] - x_bar, 2) + pow(xs[3] - x_bar, 2) + pow(xs[4] - x_bar, 2)) / xs.size());
}
int main()
{
std::cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << std::endl;
Values xs;
std::cin >> xs[0] >> xs[1] >> xs[2] >> xs[3] >> xs[4];
auto x_bar = mean(xs);
auto sigma = distribution(xs, x_bar);
std::cout << "n Mean: " << x_bar << "n Deviation: " << sigma;
}
But notice that each of those functions is doing the same operation to each of the values, then performing some final calculation based on how many. That can better be expressed using functions from <algorithm>. We then don't need to know how many elements there are.
#include <algorithm>
#include <vector>
using Values = std::vector<double>; // Any number of values
double mean(Values xs)
{
// The default combiner is +
return std::accumulate(xs.begin(), xs.end(), 0) / xs.size();
}
double distribution(Values xs, double x_bar)
{
auto sum_squares = [x_bar](double cumulative, double x){ return cumulative + pow(x - x_bar, 2); }
return sqrt(std::accumulate(xs.begin(), xs.end(), 0, sum_squares) / xs.size());
}
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not passValuesbyconstreference instead of copying it with the function call.
– Everyone
Nov 20 '18 at 12:10
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53385956%2fhow-to-pass-multiple-variables-from-main-function-down-to-other-functions-c%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have two problems of understanding basics in C++:
- Variable scope.
- Function parameters.
To briefly explain:
Variables in C++ live only in the scope they were declared in. In your case, you declared: x1, x2, x3, x4, x5 inside of the function main which means they do not exist outside that scope. The variables x1, x2, x3, x4, x5 in meanfn and deviationfn are completely different variables. They only share the same names and that's about it. Therefore, when you declared them inside the functions, they had no previously assigned values which means you cannot use them.
As for function parameters, you have to declare what parameters a function accepts in its signature. Your meanfn signature is double meanfn() which accepts no parameters. If you want it to accept 5 and only 5 double variables, you should change it to accommodate.
There are two solutions based on the explanations above:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
// Variable declarations:
double x1, x2, x3, x4, x5; // outside of 'main' making them global and everything has access to these variables.
int main()
{
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// CALL YOUR FUNCTIONS, THEY WONT BE CALLED BY THEIR OWN
double mean = meanfn();
double deviation = deviationfn(mean);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn()
{
double mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
In the solution above, the variables x1, x2, x3, x4, x5 were moved to be global variables making their access global and hence all functions can use them.
The second solution would consider parameter passing for the functions, such as:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn(double x1, double x2, double x3, double x4, double x5);
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5);
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// see the difference in the function calls?
double mean = meanfn(x1, x2, x3, x4, x5);
double deviation = deviationfn(mean, x1, x2, x3, x4, x5);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn(double x1, double x2, double x3, double x4, double x5)
{
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
add a comment |
You have two problems of understanding basics in C++:
- Variable scope.
- Function parameters.
To briefly explain:
Variables in C++ live only in the scope they were declared in. In your case, you declared: x1, x2, x3, x4, x5 inside of the function main which means they do not exist outside that scope. The variables x1, x2, x3, x4, x5 in meanfn and deviationfn are completely different variables. They only share the same names and that's about it. Therefore, when you declared them inside the functions, they had no previously assigned values which means you cannot use them.
As for function parameters, you have to declare what parameters a function accepts in its signature. Your meanfn signature is double meanfn() which accepts no parameters. If you want it to accept 5 and only 5 double variables, you should change it to accommodate.
There are two solutions based on the explanations above:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
// Variable declarations:
double x1, x2, x3, x4, x5; // outside of 'main' making them global and everything has access to these variables.
int main()
{
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// CALL YOUR FUNCTIONS, THEY WONT BE CALLED BY THEIR OWN
double mean = meanfn();
double deviation = deviationfn(mean);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn()
{
double mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
In the solution above, the variables x1, x2, x3, x4, x5 were moved to be global variables making their access global and hence all functions can use them.
The second solution would consider parameter passing for the functions, such as:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn(double x1, double x2, double x3, double x4, double x5);
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5);
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// see the difference in the function calls?
double mean = meanfn(x1, x2, x3, x4, x5);
double deviation = deviationfn(mean, x1, x2, x3, x4, x5);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn(double x1, double x2, double x3, double x4, double x5)
{
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
add a comment |
You have two problems of understanding basics in C++:
- Variable scope.
- Function parameters.
To briefly explain:
Variables in C++ live only in the scope they were declared in. In your case, you declared: x1, x2, x3, x4, x5 inside of the function main which means they do not exist outside that scope. The variables x1, x2, x3, x4, x5 in meanfn and deviationfn are completely different variables. They only share the same names and that's about it. Therefore, when you declared them inside the functions, they had no previously assigned values which means you cannot use them.
As for function parameters, you have to declare what parameters a function accepts in its signature. Your meanfn signature is double meanfn() which accepts no parameters. If you want it to accept 5 and only 5 double variables, you should change it to accommodate.
There are two solutions based on the explanations above:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
// Variable declarations:
double x1, x2, x3, x4, x5; // outside of 'main' making them global and everything has access to these variables.
int main()
{
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// CALL YOUR FUNCTIONS, THEY WONT BE CALLED BY THEIR OWN
double mean = meanfn();
double deviation = deviationfn(mean);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn()
{
double mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
In the solution above, the variables x1, x2, x3, x4, x5 were moved to be global variables making their access global and hence all functions can use them.
The second solution would consider parameter passing for the functions, such as:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn(double x1, double x2, double x3, double x4, double x5);
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5);
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// see the difference in the function calls?
double mean = meanfn(x1, x2, x3, x4, x5);
double deviation = deviationfn(mean, x1, x2, x3, x4, x5);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn(double x1, double x2, double x3, double x4, double x5)
{
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
You have two problems of understanding basics in C++:
- Variable scope.
- Function parameters.
To briefly explain:
Variables in C++ live only in the scope they were declared in. In your case, you declared: x1, x2, x3, x4, x5 inside of the function main which means they do not exist outside that scope. The variables x1, x2, x3, x4, x5 in meanfn and deviationfn are completely different variables. They only share the same names and that's about it. Therefore, when you declared them inside the functions, they had no previously assigned values which means you cannot use them.
As for function parameters, you have to declare what parameters a function accepts in its signature. Your meanfn signature is double meanfn() which accepts no parameters. If you want it to accept 5 and only 5 double variables, you should change it to accommodate.
There are two solutions based on the explanations above:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn();
double deviationfn();
// Main program:
// Variable declarations:
double x1, x2, x3, x4, x5; // outside of 'main' making them global and everything has access to these variables.
int main()
{
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// CALL YOUR FUNCTIONS, THEY WONT BE CALLED BY THEIR OWN
double mean = meanfn();
double deviation = deviationfn(mean);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn()
{
double mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
In the solution above, the variables x1, x2, x3, x4, x5 were moved to be global variables making their access global and hence all functions can use them.
The second solution would consider parameter passing for the functions, such as:
#include <iostream>
#include <cmath>
using namespace std;
// Named constant definitions (and function declarations):
//Prototype
double meanfn(double x1, double x2, double x3, double x4, double x5);
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5);
// Main program:
int main()
{
// Variable declarations:
double x1, x2, x3, x4, x5;
// Function body:
cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << endl;
cin >> x1 >> x2 >> x3 >> x4 >> x5;
// see the difference in the function calls?
double mean = meanfn(x1, x2, x3, x4, x5);
double deviation = deviationfn(mean, x1, x2, x3, x4, x5);
// print results
std::cout << "n Mean: " << mean << "n Deviation: " << deviation;
return 0;
} // end function main
double meanfn(double x1, double x2, double x3, double x4, double x5)
{
double mean;
mean = (x1 + x2 + x3 + x4 + x5) / 5;
return mean;
}// end function meanfn
double deviationfn(double mean, double x1, double x2, double x3, double x4, double x5)
{
double x,deviation;
x = mean;
deviation = sqrt((pow(x1 - x, 2) + pow(x2 - x, 2) + pow(x3 - x, 2) + pow(x4 - x, 2) + pow(x5 - x, 2)) / 5);
return deviation;
}
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
edited Nov 20 '18 at 9:06
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
answered Nov 20 '18 at 4:49
EveryoneEveryone
819318
819318
add a comment |
add a comment |
Rather than having 5 variables with numbered names, you can instead have one variable that contains all 5.
#include <iostream>
#include <cmath>
#include <array>
using Values = std::array<double, 5>;
double mean(Values xs)
{
return (xs[0] + xs[1] + xs[2] + xs[3] + xs[4]) / xs.size();
}
double distribution(Values xs, double x_bar)
{
return sqrt((pow(xs[0] - x_bar, 2) + pow(xs[1] - x_bar, 2) + pow(xs[2] - x_bar, 2) + pow(xs[3] - x_bar, 2) + pow(xs[4] - x_bar, 2)) / xs.size());
}
int main()
{
std::cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << std::endl;
Values xs;
std::cin >> xs[0] >> xs[1] >> xs[2] >> xs[3] >> xs[4];
auto x_bar = mean(xs);
auto sigma = distribution(xs, x_bar);
std::cout << "n Mean: " << x_bar << "n Deviation: " << sigma;
}
But notice that each of those functions is doing the same operation to each of the values, then performing some final calculation based on how many. That can better be expressed using functions from <algorithm>. We then don't need to know how many elements there are.
#include <algorithm>
#include <vector>
using Values = std::vector<double>; // Any number of values
double mean(Values xs)
{
// The default combiner is +
return std::accumulate(xs.begin(), xs.end(), 0) / xs.size();
}
double distribution(Values xs, double x_bar)
{
auto sum_squares = [x_bar](double cumulative, double x){ return cumulative + pow(x - x_bar, 2); }
return sqrt(std::accumulate(xs.begin(), xs.end(), 0, sum_squares) / xs.size());
}
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not passValuesbyconstreference instead of copying it with the function call.
– Everyone
Nov 20 '18 at 12:10
add a comment |
Rather than having 5 variables with numbered names, you can instead have one variable that contains all 5.
#include <iostream>
#include <cmath>
#include <array>
using Values = std::array<double, 5>;
double mean(Values xs)
{
return (xs[0] + xs[1] + xs[2] + xs[3] + xs[4]) / xs.size();
}
double distribution(Values xs, double x_bar)
{
return sqrt((pow(xs[0] - x_bar, 2) + pow(xs[1] - x_bar, 2) + pow(xs[2] - x_bar, 2) + pow(xs[3] - x_bar, 2) + pow(xs[4] - x_bar, 2)) / xs.size());
}
int main()
{
std::cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << std::endl;
Values xs;
std::cin >> xs[0] >> xs[1] >> xs[2] >> xs[3] >> xs[4];
auto x_bar = mean(xs);
auto sigma = distribution(xs, x_bar);
std::cout << "n Mean: " << x_bar << "n Deviation: " << sigma;
}
But notice that each of those functions is doing the same operation to each of the values, then performing some final calculation based on how many. That can better be expressed using functions from <algorithm>. We then don't need to know how many elements there are.
#include <algorithm>
#include <vector>
using Values = std::vector<double>; // Any number of values
double mean(Values xs)
{
// The default combiner is +
return std::accumulate(xs.begin(), xs.end(), 0) / xs.size();
}
double distribution(Values xs, double x_bar)
{
auto sum_squares = [x_bar](double cumulative, double x){ return cumulative + pow(x - x_bar, 2); }
return sqrt(std::accumulate(xs.begin(), xs.end(), 0, sum_squares) / xs.size());
}
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not passValuesbyconstreference instead of copying it with the function call.
– Everyone
Nov 20 '18 at 12:10
add a comment |
Rather than having 5 variables with numbered names, you can instead have one variable that contains all 5.
#include <iostream>
#include <cmath>
#include <array>
using Values = std::array<double, 5>;
double mean(Values xs)
{
return (xs[0] + xs[1] + xs[2] + xs[3] + xs[4]) / xs.size();
}
double distribution(Values xs, double x_bar)
{
return sqrt((pow(xs[0] - x_bar, 2) + pow(xs[1] - x_bar, 2) + pow(xs[2] - x_bar, 2) + pow(xs[3] - x_bar, 2) + pow(xs[4] - x_bar, 2)) / xs.size());
}
int main()
{
std::cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << std::endl;
Values xs;
std::cin >> xs[0] >> xs[1] >> xs[2] >> xs[3] >> xs[4];
auto x_bar = mean(xs);
auto sigma = distribution(xs, x_bar);
std::cout << "n Mean: " << x_bar << "n Deviation: " << sigma;
}
But notice that each of those functions is doing the same operation to each of the values, then performing some final calculation based on how many. That can better be expressed using functions from <algorithm>. We then don't need to know how many elements there are.
#include <algorithm>
#include <vector>
using Values = std::vector<double>; // Any number of values
double mean(Values xs)
{
// The default combiner is +
return std::accumulate(xs.begin(), xs.end(), 0) / xs.size();
}
double distribution(Values xs, double x_bar)
{
auto sum_squares = [x_bar](double cumulative, double x){ return cumulative + pow(x - x_bar, 2); }
return sqrt(std::accumulate(xs.begin(), xs.end(), 0, sum_squares) / xs.size());
}
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
Rather than having 5 variables with numbered names, you can instead have one variable that contains all 5.
#include <iostream>
#include <cmath>
#include <array>
using Values = std::array<double, 5>;
double mean(Values xs)
{
return (xs[0] + xs[1] + xs[2] + xs[3] + xs[4]) / xs.size();
}
double distribution(Values xs, double x_bar)
{
return sqrt((pow(xs[0] - x_bar, 2) + pow(xs[1] - x_bar, 2) + pow(xs[2] - x_bar, 2) + pow(xs[3] - x_bar, 2) + pow(xs[4] - x_bar, 2)) / xs.size());
}
int main()
{
std::cout << "Enter a number followed by space 5 times to calculate the mean of five numbers. " << std::endl;
Values xs;
std::cin >> xs[0] >> xs[1] >> xs[2] >> xs[3] >> xs[4];
auto x_bar = mean(xs);
auto sigma = distribution(xs, x_bar);
std::cout << "n Mean: " << x_bar << "n Deviation: " << sigma;
}
But notice that each of those functions is doing the same operation to each of the values, then performing some final calculation based on how many. That can better be expressed using functions from <algorithm>. We then don't need to know how many elements there are.
#include <algorithm>
#include <vector>
using Values = std::vector<double>; // Any number of values
double mean(Values xs)
{
// The default combiner is +
return std::accumulate(xs.begin(), xs.end(), 0) / xs.size();
}
double distribution(Values xs, double x_bar)
{
auto sum_squares = [x_bar](double cumulative, double x){ return cumulative + pow(x - x_bar, 2); }
return sqrt(std::accumulate(xs.begin(), xs.end(), 0, sum_squares) / xs.size());
}
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
edited Nov 20 '18 at 9:57
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
answered Nov 20 '18 at 9:45
CalethCaleth
17.1k22139
17.1k22139
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not passValuesbyconstreference instead of copying it with the function call.
– Everyone
Nov 20 '18 at 12:10
add a comment |
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not passValuesbyconstreference instead of copying it with the function call.
– Everyone
Nov 20 '18 at 12:10
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
This is a very good answer and I had that in mind before I typed mine. However, since the question showed lack of basic knowledge in the language, I thought this could be intimidating and would freak him out. Nonetheless, +1
– Everyone
Nov 20 '18 at 12:05
Anyway, for the sake of completeness, why not pass
Values by const reference instead of copying it with the function call.– Everyone
Nov 20 '18 at 12:10
Anyway, for the sake of completeness, why not pass
Values by const reference instead of copying it with the function call.– Everyone
Nov 20 '18 at 12:10
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53385956%2fhow-to-pass-multiple-variables-from-main-function-down-to-other-functions-c%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
Seems like the problem is with your understanding in basic C++ function calls. Since it seems like you are trying to do an assignment, Instead of posting an answer, I would recommend you to learn some C++ basics first.
– Teshan Shanuka J
Nov 20 '18 at 4:00
You have asked multiple school homwork questions here. Why don't you discuss them first with your collegues before asking "professionals" who will give you immediate solutions. Will be much better for your programming future!
– A.R.C.
Nov 20 '18 at 9:54