Mixing
Can't figure out how to give user all the options for starting, stopping, and restarting program?
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Ok so I am trying to build this random number teller which basically tells the users whether the number they input is less than, greater than, or equal to 50 and also give them the options to start, stop, and restart the "random number teller" Here is the code:
#include <iostream>
using namespace std;
main() {
cin >> boolalpha;
int invalid_answer {0};
const int const_num {50};
int random_num {};
char answer {};
int keep_going {};
while (keep_going == 0) {
while (invalid_answer == 0) {
//=======================================================================================================================================
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "Want to try again? Type "Y" or "N"";
cin >> answer;
//=======================================================================================================================================
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
//=======================================================================================================================================
while(answer == 'Y') {
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "nWant to try again? Type "Y" or "N"";
cin >> answer;
}
//=======================================================================================================================================
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
//=======================================================================================================================================
while (invalid_answer == 1) {
cout << "I'm sorry what? Please note that answers are case sensitive. Answer again: ";
cin >> answer;
if (answer == 'Y') {
invalid_answer = 0;
}
else if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
}
}
}
}
Whenever I say "N" for No I don't want to redo the random number checker, it doesn't change keep_going to 1 it just moves on to one of the other if or while statements below it. So when you input "N" it just outputs either "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " or "I'm sorry what? Please note that answers are case sensitive. Answer again: "
c++
asked Jan 3 at 9:10
ParkerParker
83
add a comment |
Ok so I am trying to build this random number teller which basically tells the users whether the number they input is less than, greater than, or equal to 50 and also give them the options to start, stop, and restart the "random number teller" Here is the code:
#include <iostream>
using namespace std;
main() {
cin >> boolalpha;
int invalid_answer {0};
const int const_num {50};
int random_num {};
char answer {};
int keep_going {};
while (keep_going == 0) {
while (invalid_answer == 0) {
//=======================================================================================================================================
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "Want to try again? Type "Y" or "N"";
cin >> answer;
//=======================================================================================================================================
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
//=======================================================================================================================================
while(answer == 'Y') {
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "nWant to try again? Type "Y" or "N"";
cin >> answer;
}
//=======================================================================================================================================
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
//=======================================================================================================================================
while (invalid_answer == 1) {
cout << "I'm sorry what? Please note that answers are case sensitive. Answer again: ";
cin >> answer;
if (answer == 'Y') {
invalid_answer = 0;
}
else if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
}
}
}
}
Whenever I say "N" for No I don't want to redo the random number checker, it doesn't change keep_going to 1 it just moves on to one of the other if or while statements below it. So when you input "N" it just outputs either "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " or "I'm sorry what? Please note that answers are case sensitive. Answer again: "
c++
asked Jan 3 at 9:10
ParkerParker
83
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the conditionanswer != 'Y' || answer != 'N', and what happens ifansweris equal to'N'.
– Some programmer dude
Jan 3 at 9:15
add a comment |
Ok so I am trying to build this random number teller which basically tells the users whether the number they input is less than, greater than, or equal to 50 and also give them the options to start, stop, and restart the "random number teller" Here is the code:
#include <iostream>
using namespace std;
main() {
cin >> boolalpha;
int invalid_answer {0};
const int const_num {50};
int random_num {};
char answer {};
int keep_going {};
while (keep_going == 0) {
while (invalid_answer == 0) {
//=======================================================================================================================================
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "Want to try again? Type "Y" or "N"";
cin >> answer;
//=======================================================================================================================================
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
//=======================================================================================================================================
while(answer == 'Y') {
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "nWant to try again? Type "Y" or "N"";
cin >> answer;
}
//=======================================================================================================================================
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
//=======================================================================================================================================
while (invalid_answer == 1) {
cout << "I'm sorry what? Please note that answers are case sensitive. Answer again: ";
cin >> answer;
if (answer == 'Y') {
invalid_answer = 0;
}
else if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
}
}
}
}
Whenever I say "N" for No I don't want to redo the random number checker, it doesn't change keep_going to 1 it just moves on to one of the other if or while statements below it. So when you input "N" it just outputs either "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " or "I'm sorry what? Please note that answers are case sensitive. Answer again: "
c++
asked Jan 3 at 9:10
ParkerParker
83
Ok so I am trying to build this random number teller which basically tells the users whether the number they input is less than, greater than, or equal to 50 and also give them the options to start, stop, and restart the "random number teller" Here is the code:
#include <iostream>
using namespace std;
main() {
cin >> boolalpha;
int invalid_answer {0};
const int const_num {50};
int random_num {};
char answer {};
int keep_going {};
while (keep_going == 0) {
while (invalid_answer == 0) {
//=======================================================================================================================================
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "Want to try again? Type "Y" or "N"";
cin >> answer;
//=======================================================================================================================================
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
//=======================================================================================================================================
while(answer == 'Y') {
cout << "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " << endl;
cin >> random_num;
if (random_num > const_num) {
cout << random_num << " is greater than " << const_num;
}
else if (random_num == const_num) {
cout << random_num << " is the same as " << const_num << endl;
}
else {
cout << random_num << " is less than " << const_num << endl;
}
cout << "nWant to try again? Type "Y" or "N"";
cin >> answer;
}
//=======================================================================================================================================
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
//=======================================================================================================================================
while (invalid_answer == 1) {
cout << "I'm sorry what? Please note that answers are case sensitive. Answer again: ";
cin >> answer;
if (answer == 'Y') {
invalid_answer = 0;
}
else if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}
}
}
}
}
Whenever I say "N" for No I don't want to redo the random number checker, it doesn't change keep_going to 1 it just moves on to one of the other if or while statements below it. So when you input "N" it just outputs either "Enter a random number and we will tell you if it is greater than or less than " << const_num << ": " or "I'm sorry what? Please note that answers are case sensitive. Answer again: "
c++
c++
asked Jan 3 at 9:10
ParkerParker
83
asked Jan 3 at 9:10
ParkerParker
83
asked Jan 3 at 9:10
ParkerParker
83
asked Jan 3 at 9:10
ParkerParker
83
asked Jan 3 at 9:10
ParkerParker
83
83
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the conditionanswer != 'Y' || answer != 'N', and what happens ifansweris equal to'N'.
– Some programmer dude
Jan 3 at 9:15
add a comment |
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the conditionanswer != 'Y' || answer != 'N', and what happens ifansweris equal to'N'.
– Some programmer dude
Jan 3 at 9:15
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the condition
answer != 'Y' || answer != 'N', and what happens if answer is equal to 'N'.– Some programmer dude
Jan 3 at 9:15
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the condition
answer != 'Y' || answer != 'N', and what happens if answer is equal to 'N'.– Some programmer dude
Jan 3 at 9:15
add a comment |
2 Answers
2
active
oldest
votes
The problem is with this bit of code:
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
When answer is 'N', answer != 'Y' is true and invalid_answer is set to 1 (because of short-circuit evaluation the rhs of the logical OR is not even evaluated - see quote below).
So the execution will enter the while
while (invalid_answer == 1)
and will print the statements.
You can correct this by:
if (answer == 'Y' || answer == 'N') { //if input is either 'Y' or 'N'
invalid_answer = 0;
}
else { //for all other inputs
invalid_answer = 1;
}
Builtin operators
&&and||perform short-circuit evaluation (do not evaluate the second operand if the result is known after evaluating the first), but overloaded operators behave like regular function calls and always evaluate both operands
Also note that main should have the type int.
answered Jan 3 at 9:19
P.WP.W
18.4k41758
Short-circuit evaluation has no effect on the result here. Every character is either notYor notN.
– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', theifwould fail and theinvalid_answerwould not be 1. I will update the answer and clarify.
– P.W
Jan 3 at 9:36
add a comment |
I figured it out right after I posted the question haha, basically the answer above was correct so I had to split that if statement into 2 others, in which I added an else statement to each also that said invalid_answer = 0; to make sure. But then after the user's second time using the program, if they wanted to quit it wouldn't let them and would just restart it again. I solved that by adding
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}`
to the bottom of the while(answer == 'Y') loop.
answered Jan 3 at 9:51
ParkerParker
83
add a comment |
Your Answer
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The problem is with this bit of code:
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
When answer is 'N', answer != 'Y' is true and invalid_answer is set to 1 (because of short-circuit evaluation the rhs of the logical OR is not even evaluated - see quote below).
So the execution will enter the while
while (invalid_answer == 1)
and will print the statements.
You can correct this by:
if (answer == 'Y' || answer == 'N') { //if input is either 'Y' or 'N'
invalid_answer = 0;
}
else { //for all other inputs
invalid_answer = 1;
}
Builtin operators
&&and||perform short-circuit evaluation (do not evaluate the second operand if the result is known after evaluating the first), but overloaded operators behave like regular function calls and always evaluate both operands
Also note that main should have the type int.
answered Jan 3 at 9:19
P.WP.W
18.4k41758
Short-circuit evaluation has no effect on the result here. Every character is either notYor notN.
– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', theifwould fail and theinvalid_answerwould not be 1. I will update the answer and clarify.
– P.W
Jan 3 at 9:36
add a comment |
The problem is with this bit of code:
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
When answer is 'N', answer != 'Y' is true and invalid_answer is set to 1 (because of short-circuit evaluation the rhs of the logical OR is not even evaluated - see quote below).
So the execution will enter the while
while (invalid_answer == 1)
and will print the statements.
You can correct this by:
if (answer == 'Y' || answer == 'N') { //if input is either 'Y' or 'N'
invalid_answer = 0;
}
else { //for all other inputs
invalid_answer = 1;
}
Builtin operators
&&and||perform short-circuit evaluation (do not evaluate the second operand if the result is known after evaluating the first), but overloaded operators behave like regular function calls and always evaluate both operands
Also note that main should have the type int.
answered Jan 3 at 9:19
P.WP.W
18.4k41758
Short-circuit evaluation has no effect on the result here. Every character is either notYor notN.
– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', theifwould fail and theinvalid_answerwould not be 1. I will update the answer and clarify.
– P.W
Jan 3 at 9:36
add a comment |
The problem is with this bit of code:
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
When answer is 'N', answer != 'Y' is true and invalid_answer is set to 1 (because of short-circuit evaluation the rhs of the logical OR is not even evaluated - see quote below).
So the execution will enter the while
while (invalid_answer == 1)
and will print the statements.
You can correct this by:
if (answer == 'Y' || answer == 'N') { //if input is either 'Y' or 'N'
invalid_answer = 0;
}
else { //for all other inputs
invalid_answer = 1;
}
Builtin operators
&&and||perform short-circuit evaluation (do not evaluate the second operand if the result is known after evaluating the first), but overloaded operators behave like regular function calls and always evaluate both operands
Also note that main should have the type int.
answered Jan 3 at 9:19
P.WP.W
18.4k41758
The problem is with this bit of code:
if (answer != 'Y' || answer != 'N') {
invalid_answer = 1;
}
When answer is 'N', answer != 'Y' is true and invalid_answer is set to 1 (because of short-circuit evaluation the rhs of the logical OR is not even evaluated - see quote below).
So the execution will enter the while
while (invalid_answer == 1)
and will print the statements.
You can correct this by:
if (answer == 'Y' || answer == 'N') { //if input is either 'Y' or 'N'
invalid_answer = 0;
}
else { //for all other inputs
invalid_answer = 1;
}
Builtin operators
&&and||perform short-circuit evaluation (do not evaluate the second operand if the result is known after evaluating the first), but overloaded operators behave like regular function calls and always evaluate both operands
Also note that main should have the type int.
answered Jan 3 at 9:19
P.WP.W
18.4k41758
edited Jan 3 at 10:33
answered Jan 3 at 9:19
P.WP.W
18.4k41758
answered Jan 3 at 9:19
P.WP.W
18.4k41758
answered Jan 3 at 9:19
P.WP.W
18.4k41758
18.4k41758
Short-circuit evaluation has no effect on the result here. Every character is either notYor notN.
– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', theifwould fail and theinvalid_answerwould not be 1. I will update the answer and clarify.
– P.W
Jan 3 at 9:36
add a comment |
Short-circuit evaluation has no effect on the result here. Every character is either notYor notN.
– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', theifwould fail and theinvalid_answerwould not be 1. I will update the answer and clarify.
– P.W
Jan 3 at 9:36
Short-circuit evaluation has no effect on the result here. Every character is either not
Y or not N.– David Schwartz
Jan 3 at 9:33
Short-circuit evaluation has no effect on the result here. Every character is either not
Y or not N.– David Schwartz
Jan 3 at 9:33
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', the
if would fail and the invalid_answer would not be 1. I will update the answer and clarify.– P.W
Jan 3 at 9:36
Yes. The OP was asking for the specific case of 'N'. He must have thought that if input is 'N', the
if would fail and the invalid_answer would not be 1. I will update the answer and clarify.– P.W
Jan 3 at 9:36
add a comment |
I figured it out right after I posted the question haha, basically the answer above was correct so I had to split that if statement into 2 others, in which I added an else statement to each also that said invalid_answer = 0; to make sure. But then after the user's second time using the program, if they wanted to quit it wouldn't let them and would just restart it again. I solved that by adding
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}`
to the bottom of the while(answer == 'Y') loop.
answered Jan 3 at 9:51
ParkerParker
83
add a comment |
I figured it out right after I posted the question haha, basically the answer above was correct so I had to split that if statement into 2 others, in which I added an else statement to each also that said invalid_answer = 0; to make sure. But then after the user's second time using the program, if they wanted to quit it wouldn't let them and would just restart it again. I solved that by adding
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}`
to the bottom of the while(answer == 'Y') loop.
answered Jan 3 at 9:51
ParkerParker
83
add a comment |
I figured it out right after I posted the question haha, basically the answer above was correct so I had to split that if statement into 2 others, in which I added an else statement to each also that said invalid_answer = 0; to make sure. But then after the user's second time using the program, if they wanted to quit it wouldn't let them and would just restart it again. I solved that by adding
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}`
to the bottom of the while(answer == 'Y') loop.
answered Jan 3 at 9:51
ParkerParker
83
I figured it out right after I posted the question haha, basically the answer above was correct so I had to split that if statement into 2 others, in which I added an else statement to each also that said invalid_answer = 0; to make sure. But then after the user's second time using the program, if they wanted to quit it wouldn't let them and would just restart it again. I solved that by adding
if (answer == 'N') {
cout << "Ok then, sorry to see you miss out" << endl;
keep_going = 1;
}`
to the bottom of the while(answer == 'Y') loop.
answered Jan 3 at 9:51
ParkerParker
83
answered Jan 3 at 9:51
ParkerParker
83
answered Jan 3 at 9:51
ParkerParker
83
answered Jan 3 at 9:51
ParkerParker
83
83
add a comment |
add a comment |
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Required, but never shown
Remember that C++ uses short-circuit evaluation for logical expression. With that information, think a little bit about the condition
answer != 'Y' || answer != 'N', and what happens ifansweris equal to'N'.– Some programmer dude
Jan 3 at 9:15