Mixing
Is it possible to add to char type variable to get string ?
-1
Let's say I have 2 string variables
string str1 = "A";
string str2 = "B";
string str3;
Doing
str3 = str1 + str2;
Would get me
str3 = "AB"
My question is would it be possible to do addition in a somewhat similar fashion if instead of string type, str1 and str2, there would be char type values? If not, could anyone suggest me some smart workarounds if possible?
char str1 = 'A';
char str2 = 'B';
string str3; // <------ I need to get "AB" in str3 somehow
Thank you
c++ string char
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
add a comment |
-1
Let's say I have 2 string variables
string str1 = "A";
string str2 = "B";
string str3;
Doing
str3 = str1 + str2;
Would get me
str3 = "AB"
My question is would it be possible to do addition in a somewhat similar fashion if instead of string type, str1 and str2, there would be char type values? If not, could anyone suggest me some smart workarounds if possible?
char str1 = 'A';
char str2 = 'B';
string str3; // <------ I need to get "AB" in str3 somehow
Thank you
c++ string char
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
@Jokubas11.. see more ways given below..
– code_cody97
Nov 22 '18 at 9:36
add a comment |
-1
-1
-1
Let's say I have 2 string variables
string str1 = "A";
string str2 = "B";
string str3;
Doing
str3 = str1 + str2;
Would get me
str3 = "AB"
My question is would it be possible to do addition in a somewhat similar fashion if instead of string type, str1 and str2, there would be char type values? If not, could anyone suggest me some smart workarounds if possible?
char str1 = 'A';
char str2 = 'B';
string str3; // <------ I need to get "AB" in str3 somehow
Thank you
c++ string char
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
Let's say I have 2 string variables
string str1 = "A";
string str2 = "B";
string str3;
Doing
str3 = str1 + str2;
Would get me
str3 = "AB"
My question is would it be possible to do addition in a somewhat similar fashion if instead of string type, str1 and str2, there would be char type values? If not, could anyone suggest me some smart workarounds if possible?
char str1 = 'A';
char str2 = 'B';
string str3; // <------ I need to get "AB" in str3 somehow
Thank you
c++ string char
c++ string char
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
edited Nov 19 '18 at 15:24
eyllanesc
74.2k103156
74.2k103156
asked Nov 19 '18 at 15:12
Jokubas11
1
asked Nov 19 '18 at 15:12
Jokubas11
1
asked Nov 19 '18 at 15:12
Jokubas11
1
1
See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
@Jokubas11.. see more ways given below..
– code_cody97
Nov 22 '18 at 9:36
add a comment |
See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
@Jokubas11.. see more ways given below..
– code_cody97
Nov 22 '18 at 9:36
See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
@Jokubas11 .. see more ways given below..– code_cody97
Nov 22 '18 at 9:36
@Jokubas11 .. see more ways given below..– code_cody97
Nov 22 '18 at 9:36
add a comment |
4 Answers
4
active
oldest
votes
4
Sure. There are several ways. Heres a few.
#include<string>
#include<iostream>
using namespace std;
int main()
{
char str1 = 'A';
char str2 = 'B';
string str3;
//using operator+=
str3 += str1;
str3 += str2;
cout << str3 << endl;
str3.clear();
//using push_back
str3.push_back(str1);
str3.push_back(str2);
cout << str3 << endl;
str3.clear();
//using array access
str3.resize(2);
str3[0] = str1;
str3[1] = str2;
cout << str3 << endl;
return 0;
}
answered Nov 19 '18 at 15:21
johnathan
2,158819
add a comment |
3
Yes, it's entirely possible to concatenate char to a std::string, you just need one of the operands to be a std::string, otherwise you are essentially adding integers.
#include <iostream>
int main()
{
char a = 'A';
char b = 'B';
std::string str = std::string() + a + b;
return 0;
}
answered Nov 19 '18 at 15:22
not an alien
215110
add a comment |
1
It is possible to create the string in one go. This is the most efficient way also:
char a = 'A';
char b = 'B';
std::string str{a, b};
std::cout << str << std::endl; // "AB"
This uses the initializer_list constructor of std::string.
For a more general way of constructing a string (from any type that works with basic_ostream::operator<<) you can use stringstream:
char a = 'A';
char b = 'B';
std::stringstream ss;
ss << a << b;
std::string str2 = ss.str();
std::cout << str2 << std::endl; // "AB"
answered Nov 19 '18 at 15:28
bolov
30.7k668128
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenatestd::strings andchars usingoperator+(), but after re-reading the question I am not so certain. My mistake.
– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.std::string::operator+,std::string::push_back,std::ostream::operator<<etc. The fact that the op was familiar with one waystd::string::operator+doesn't mean we need to limit ourselves to just finding anoperator+.
– bolov
Nov 19 '18 at 16:23
add a comment |
-1
There are following ways..
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
int main()
{
char a = 'A';
char b = 'B';
//using string class fill constructor
string str3;
str3 = string(1,a) + string(1,b);
cout << str3;
str3.clear();
cout << endl;
//using stringstream
stringstream ss;
ss << a << b;
str3 = ss.str();
cout << str3;
str3.clear();
cout << endl;
//using string push_back
str3.pb(a);
str3.pb(b);
cout << str3;
str3.clear();
cout << endl;
//using string::operator+=
str3+=a;
str3+=b;
cout << str3;
str3.clear();
cout << endl;
//using string::append
str3.append(1,a);
str3.append(1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::insert
str3.insert(0,1,a);
str3.insert(1,1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::replace
str3.replace(0,1,1,a);
str3.replace(1,1,1,b);
cout << str3;
str3.clear();
cout << endl;
return 0;
}
answered Nov 19 '18 at 15:28
code_cody97
709
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
Sure. There are several ways. Heres a few.
#include<string>
#include<iostream>
using namespace std;
int main()
{
char str1 = 'A';
char str2 = 'B';
string str3;
//using operator+=
str3 += str1;
str3 += str2;
cout << str3 << endl;
str3.clear();
//using push_back
str3.push_back(str1);
str3.push_back(str2);
cout << str3 << endl;
str3.clear();
//using array access
str3.resize(2);
str3[0] = str1;
str3[1] = str2;
cout << str3 << endl;
return 0;
}
answered Nov 19 '18 at 15:21
johnathan
2,158819
add a comment |
4
Sure. There are several ways. Heres a few.
#include<string>
#include<iostream>
using namespace std;
int main()
{
char str1 = 'A';
char str2 = 'B';
string str3;
//using operator+=
str3 += str1;
str3 += str2;
cout << str3 << endl;
str3.clear();
//using push_back
str3.push_back(str1);
str3.push_back(str2);
cout << str3 << endl;
str3.clear();
//using array access
str3.resize(2);
str3[0] = str1;
str3[1] = str2;
cout << str3 << endl;
return 0;
}
answered Nov 19 '18 at 15:21
johnathan
2,158819
add a comment |
4
4
4
Sure. There are several ways. Heres a few.
#include<string>
#include<iostream>
using namespace std;
int main()
{
char str1 = 'A';
char str2 = 'B';
string str3;
//using operator+=
str3 += str1;
str3 += str2;
cout << str3 << endl;
str3.clear();
//using push_back
str3.push_back(str1);
str3.push_back(str2);
cout << str3 << endl;
str3.clear();
//using array access
str3.resize(2);
str3[0] = str1;
str3[1] = str2;
cout << str3 << endl;
return 0;
}
answered Nov 19 '18 at 15:21
johnathan
2,158819
Sure. There are several ways. Heres a few.
#include<string>
#include<iostream>
using namespace std;
int main()
{
char str1 = 'A';
char str2 = 'B';
string str3;
//using operator+=
str3 += str1;
str3 += str2;
cout << str3 << endl;
str3.clear();
//using push_back
str3.push_back(str1);
str3.push_back(str2);
cout << str3 << endl;
str3.clear();
//using array access
str3.resize(2);
str3[0] = str1;
str3[1] = str2;
cout << str3 << endl;
return 0;
}
answered Nov 19 '18 at 15:21
johnathan
2,158819
edited Nov 19 '18 at 16:38
answered Nov 19 '18 at 15:21
johnathan
2,158819
answered Nov 19 '18 at 15:21
johnathan
2,158819
answered Nov 19 '18 at 15:21
johnathan
2,158819
2,158819
add a comment |
add a comment |
3
Yes, it's entirely possible to concatenate char to a std::string, you just need one of the operands to be a std::string, otherwise you are essentially adding integers.
#include <iostream>
int main()
{
char a = 'A';
char b = 'B';
std::string str = std::string() + a + b;
return 0;
}
answered Nov 19 '18 at 15:22
not an alien
215110
add a comment |
3
Yes, it's entirely possible to concatenate char to a std::string, you just need one of the operands to be a std::string, otherwise you are essentially adding integers.
#include <iostream>
int main()
{
char a = 'A';
char b = 'B';
std::string str = std::string() + a + b;
return 0;
}
answered Nov 19 '18 at 15:22
not an alien
215110
add a comment |
3
3
3
Yes, it's entirely possible to concatenate char to a std::string, you just need one of the operands to be a std::string, otherwise you are essentially adding integers.
#include <iostream>
int main()
{
char a = 'A';
char b = 'B';
std::string str = std::string() + a + b;
return 0;
}
answered Nov 19 '18 at 15:22
not an alien
215110
Yes, it's entirely possible to concatenate char to a std::string, you just need one of the operands to be a std::string, otherwise you are essentially adding integers.
#include <iostream>
int main()
{
char a = 'A';
char b = 'B';
std::string str = std::string() + a + b;
return 0;
}
answered Nov 19 '18 at 15:22
not an alien
215110
answered Nov 19 '18 at 15:22
not an alien
215110
answered Nov 19 '18 at 15:22
not an alien
215110
answered Nov 19 '18 at 15:22
not an alien
215110
215110
add a comment |
add a comment |
1
It is possible to create the string in one go. This is the most efficient way also:
char a = 'A';
char b = 'B';
std::string str{a, b};
std::cout << str << std::endl; // "AB"
This uses the initializer_list constructor of std::string.
For a more general way of constructing a string (from any type that works with basic_ostream::operator<<) you can use stringstream:
char a = 'A';
char b = 'B';
std::stringstream ss;
ss << a << b;
std::string str2 = ss.str();
std::cout << str2 << std::endl; // "AB"
answered Nov 19 '18 at 15:28
bolov
30.7k668128
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenatestd::strings andchars usingoperator+(), but after re-reading the question I am not so certain. My mistake.
– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.std::string::operator+,std::string::push_back,std::ostream::operator<<etc. The fact that the op was familiar with one waystd::string::operator+doesn't mean we need to limit ourselves to just finding anoperator+.
– bolov
Nov 19 '18 at 16:23
add a comment |
1
It is possible to create the string in one go. This is the most efficient way also:
char a = 'A';
char b = 'B';
std::string str{a, b};
std::cout << str << std::endl; // "AB"
This uses the initializer_list constructor of std::string.
For a more general way of constructing a string (from any type that works with basic_ostream::operator<<) you can use stringstream:
char a = 'A';
char b = 'B';
std::stringstream ss;
ss << a << b;
std::string str2 = ss.str();
std::cout << str2 << std::endl; // "AB"
answered Nov 19 '18 at 15:28
bolov
30.7k668128
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenatestd::strings andchars usingoperator+(), but after re-reading the question I am not so certain. My mistake.
– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.std::string::operator+,std::string::push_back,std::ostream::operator<<etc. The fact that the op was familiar with one waystd::string::operator+doesn't mean we need to limit ourselves to just finding anoperator+.
– bolov
Nov 19 '18 at 16:23
add a comment |
1
1
1
It is possible to create the string in one go. This is the most efficient way also:
char a = 'A';
char b = 'B';
std::string str{a, b};
std::cout << str << std::endl; // "AB"
This uses the initializer_list constructor of std::string.
For a more general way of constructing a string (from any type that works with basic_ostream::operator<<) you can use stringstream:
char a = 'A';
char b = 'B';
std::stringstream ss;
ss << a << b;
std::string str2 = ss.str();
std::cout << str2 << std::endl; // "AB"
answered Nov 19 '18 at 15:28
bolov
30.7k668128
It is possible to create the string in one go. This is the most efficient way also:
char a = 'A';
char b = 'B';
std::string str{a, b};
std::cout << str << std::endl; // "AB"
This uses the initializer_list constructor of std::string.
For a more general way of constructing a string (from any type that works with basic_ostream::operator<<) you can use stringstream:
char a = 'A';
char b = 'B';
std::stringstream ss;
ss << a << b;
std::string str2 = ss.str();
std::cout << str2 << std::endl; // "AB"
answered Nov 19 '18 at 15:28
bolov
30.7k668128
edited Nov 19 '18 at 15:31
answered Nov 19 '18 at 15:28
bolov
30.7k668128
answered Nov 19 '18 at 15:28
bolov
30.7k668128
answered Nov 19 '18 at 15:28
bolov
30.7k668128
30.7k668128
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenatestd::strings andchars usingoperator+(), but after re-reading the question I am not so certain. My mistake.
– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.std::string::operator+,std::string::push_back,std::ostream::operator<<etc. The fact that the op was familiar with one waystd::string::operator+doesn't mean we need to limit ourselves to just finding anoperator+.
– bolov
Nov 19 '18 at 16:23
add a comment |
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenatestd::strings andchars usingoperator+(), but after re-reading the question I am not so certain. My mistake.
– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.std::string::operator+,std::string::push_back,std::ostream::operator<<etc. The fact that the op was familiar with one waystd::string::operator+doesn't mean we need to limit ourselves to just finding anoperator+.
– bolov
Nov 19 '18 at 16:23
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
Most efficient, indeed. But perhaps not what the OP was asking for exactly.
– not an alien
Nov 19 '18 at 15:30
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien how so?
– bolov
Nov 19 '18 at 15:32
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
@notanalien "would it be possible to do addition in a somewhat similar fashion [...]? If not, could anyone suggest me some smart workarounds if possible?" how is this not what the OP was asking for? And even if the OP didn't ask for it, it's a good way to achieve this. Remember this site doesn't benefit the OP only.
– bolov
Nov 19 '18 at 15:35
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenate
std::strings and chars using operator+(), but after re-reading the question I am not so certain. My mistake.– not an alien
Nov 19 '18 at 15:40
My interpretation of "do addition in a somewhat similar fashion" was along the lines of "concatenate
std::strings and chars using operator+(), but after re-reading the question I am not so certain. My mistake.– not an alien
Nov 19 '18 at 15:40
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.
std::string::operator+, std::string::push_back, std::ostream::operator<< etc. The fact that the op was familiar with one way std::string::operator+ doesn't mean we need to limit ourselves to just finding an operator+.– bolov
Nov 19 '18 at 16:23
@notanalien the idea is indeed to concatenate chars into a string. That is the operation asked for. Now this can be achieved in C++ several ways.
std::string::operator+, std::string::push_back, std::ostream::operator<< etc. The fact that the op was familiar with one way std::string::operator+ doesn't mean we need to limit ourselves to just finding an operator+.– bolov
Nov 19 '18 at 16:23
add a comment |
-1
There are following ways..
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
int main()
{
char a = 'A';
char b = 'B';
//using string class fill constructor
string str3;
str3 = string(1,a) + string(1,b);
cout << str3;
str3.clear();
cout << endl;
//using stringstream
stringstream ss;
ss << a << b;
str3 = ss.str();
cout << str3;
str3.clear();
cout << endl;
//using string push_back
str3.pb(a);
str3.pb(b);
cout << str3;
str3.clear();
cout << endl;
//using string::operator+=
str3+=a;
str3+=b;
cout << str3;
str3.clear();
cout << endl;
//using string::append
str3.append(1,a);
str3.append(1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::insert
str3.insert(0,1,a);
str3.insert(1,1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::replace
str3.replace(0,1,1,a);
str3.replace(1,1,1,b);
cout << str3;
str3.clear();
cout << endl;
return 0;
}
answered Nov 19 '18 at 15:28
code_cody97
709
add a comment |
-1
There are following ways..
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
int main()
{
char a = 'A';
char b = 'B';
//using string class fill constructor
string str3;
str3 = string(1,a) + string(1,b);
cout << str3;
str3.clear();
cout << endl;
//using stringstream
stringstream ss;
ss << a << b;
str3 = ss.str();
cout << str3;
str3.clear();
cout << endl;
//using string push_back
str3.pb(a);
str3.pb(b);
cout << str3;
str3.clear();
cout << endl;
//using string::operator+=
str3+=a;
str3+=b;
cout << str3;
str3.clear();
cout << endl;
//using string::append
str3.append(1,a);
str3.append(1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::insert
str3.insert(0,1,a);
str3.insert(1,1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::replace
str3.replace(0,1,1,a);
str3.replace(1,1,1,b);
cout << str3;
str3.clear();
cout << endl;
return 0;
}
answered Nov 19 '18 at 15:28
code_cody97
709
add a comment |
-1
-1
-1
There are following ways..
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
int main()
{
char a = 'A';
char b = 'B';
//using string class fill constructor
string str3;
str3 = string(1,a) + string(1,b);
cout << str3;
str3.clear();
cout << endl;
//using stringstream
stringstream ss;
ss << a << b;
str3 = ss.str();
cout << str3;
str3.clear();
cout << endl;
//using string push_back
str3.pb(a);
str3.pb(b);
cout << str3;
str3.clear();
cout << endl;
//using string::operator+=
str3+=a;
str3+=b;
cout << str3;
str3.clear();
cout << endl;
//using string::append
str3.append(1,a);
str3.append(1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::insert
str3.insert(0,1,a);
str3.insert(1,1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::replace
str3.replace(0,1,1,a);
str3.replace(1,1,1,b);
cout << str3;
str3.clear();
cout << endl;
return 0;
}
answered Nov 19 '18 at 15:28
code_cody97
709
There are following ways..
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
int main()
{
char a = 'A';
char b = 'B';
//using string class fill constructor
string str3;
str3 = string(1,a) + string(1,b);
cout << str3;
str3.clear();
cout << endl;
//using stringstream
stringstream ss;
ss << a << b;
str3 = ss.str();
cout << str3;
str3.clear();
cout << endl;
//using string push_back
str3.pb(a);
str3.pb(b);
cout << str3;
str3.clear();
cout << endl;
//using string::operator+=
str3+=a;
str3+=b;
cout << str3;
str3.clear();
cout << endl;
//using string::append
str3.append(1,a);
str3.append(1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::insert
str3.insert(0,1,a);
str3.insert(1,1,b);
cout << str3;
str3.clear();
cout << endl;
//using string::replace
str3.replace(0,1,1,a);
str3.replace(1,1,1,b);
cout << str3;
str3.clear();
cout << endl;
return 0;
}
answered Nov 19 '18 at 15:28
code_cody97
709
edited Nov 19 '18 at 18:27
answered Nov 19 '18 at 15:28
code_cody97
709
answered Nov 19 '18 at 15:28
code_cody97
709
answered Nov 19 '18 at 15:28
code_cody97
709
709
add a comment |
add a comment |
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See std::string::push_back
– Ayxan
Nov 19 '18 at 15:24
@Jokubas11.. see more ways given below..– code_cody97
Nov 22 '18 at 9:36