INT to STR

[Marcel]

The small code underneath is displaying the text 'hello2' as output. I
supposed it to display hello50. I think i will have to convert the int to a
string or am i wrong? If so, what is the right command? Sorry i am a
beginner....

Thanks in advance!

#include <iostream>
#include <string>

int main() {

int red = 50;

std::string temp = "";

temp += "hello";

temp += red;

std::cout << temp;

std::cin.get();

return 0;
}

Regards,

Marcel

 

[keanu]

The small code underneath is displaying the text 'hello2' as output. I
supposed it to display hello50. I think i will have to convert the int to
a string or am i wrong? If so, what is the right command? Sorry i am a
beginner....

Thanks in advance!

#include <iostream>
#include <string>

#include <sstream>
template<typename T> std::string toStr(T var) {
std:stringstream tmp; tmp << var; return tmp.str();
}

int main() {
int red = 50;
std::string temp = "";
temp += "hello";

//temp += red;

 

[Marcel]

#include <sstream>
template<typename T> std::string toStr(T var) {
std:stringstream tmp; tmp << var; return tmp.str();
}
//temp += red;

Thanks a lot,

That's more complicated than i expected!

Marcel

 

[red floyd]

The small code underneath is displaying the text 'hello2' as output. I
supposed it to display hello50. I think i will have to convert the int to a
string or am i wrong? If so, what is the right command? Sorry i am a
beginner....

Thanks in advance!

The other poster showed what to do. I'll explain why it happened.

#include <iostream>
#include <string>

int main() {

int red = 50;

Red is of type int.

std::string temp = "";

temp += "hello";

So far you're fine.

temp += red;

There is no string:perator+=(int), however there is a string:perator+=(char).
So, what happens is that the standard conversion from int to char takes place, and operator+=(char) is called, and the character
with the value of 50 (which in ASCII is '2') is appended to your string.

std::cout << temp;

std::cin.get();

return 0;
}

I suspect that you were expecting a std::string to behave similarly to a VB String variable. Alas, it doesn't work that way, and
you need to do the string conversion yourself before using +=.