I want to convert an integer to a string. I tried it this way but this didn't work
void foo() { int y = 1; string x = static_cast<string>(y); } 
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3 Answers
The std::to_string function should do it:
string x = std::to_string(y); For the opposite, it's std::stoi:
int z = std::stoi(y, nullptr, 10); answered Sep 24, 2013 at 18:43
Some programmer dude
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5 Comments
Bathsheba
oh that's cute +1; is that a C++11 thing?
Mohammed Hossain
Note: this is only supported by compilers with C++11 support (which you should probably have, by now).
Some programmer dude
@Bathsheba Yes, it's new in C++11. For older compilers, your version is the solution to use (and most likely how
to_string works).
dcp
On MS Visual Studio 2010 C++, I get this error: error C2668: 'std::to_string' : ambiguous call to overloaded function. If I cast x to (long double), everything works ok. Actually, looks like some other folks had the same problem: stackoverflow.com/questions/14617950/… Anyway, not sure this answer will work on all modern compilers. VS 2010 isn't that old. But that's C++, what can you say :).
Pete Becker
VS 2010 is not a full implementation of C++ 2011.
No that will not work since int and std::string are not related in any class heirarchy. Therefore a static_cast will fail.
An easy way (though not necessarily the fastest way) is to write
std::stringsteam ss; ss << y; std::string x = ss.str(); But, if you have a C++11 compiler, Joachim Pileborg's solution is much better.
Comments
Can have this :
template <typename T> string CovertToString ( T Number ) { ostringstream ss; ss << Number; return ss.str(); } 
1 Comment
cdhowie
I'd consider calling it something besides
NumberToString(), since this will work for any object for which such a streaming operator exists.