How to enable the _Generic keyword - c++11

Question

I have this test source:
#include <stdio.h>
int main()
{
int x;
printf("x=%d\n", _Generic('x', int: 1, default: 0));
return 0;
}
Compiling with c++ (from GCC 4.9.2) fails:
t.cpp: In function ‘int main()’:
t.cpp:7:33: error: expected primary-expression before ‘int’
printf("x=%d\n", _Generic('x', int: 1, default: 0));
^
t.cpp:7:41: error: expected primary-expression before ‘default’
printf("x=%d\n", _Generic('x', int: 1, default: 0));
^
t.cpp:7:51: error: ‘_Generic’ was not declared in this scope
printf("x=%d\n", _Generic('x', int: 1, default: 0));
The compiler arguments are:
c++ --std=c++11 t.cpp -o t
What am I doing wrong?

Answer 1

_Generic is a C11 feature. It is not present in C++ (any version at least up to C++14 - I don't really expect it to be added either).
If you want to use it, you'll need to write C code, and use a compiler that supports that standard (reasonably recent versions of gcc and clang do for example, using -std=c11).
If you want to write C++, use overloading or templates instead, for example:
#include <iostream>
int foo(int) { return 1; }
int foo(char) { return 0; }
int main()
{
std::cout << "x=" << foo('x') << std::endl;
}
This prints x=0 in C++, the foo(char) overload is the best match.
Note that there's difference between C and C++ that might trick you here too: 'x' is a char in C++. It's an int in C. So if _Generic had been implemented (maybe as an extension) by your compiler, chances are you'd get different output when compiling your example as C versus compiling as C++.

Answer 2

Here's the C++ form (forgive me for using the using directive, I know its bad form):
#include <iostream>
using namespace std;
template< typename T> T do_something(T argument) {
// Put here what you need
}
int main()
{
int x;
cout << "x" << (x = do_something(x));
return 0;
}
_Generic is C11, you're probably using a C++ compiler when you meant to use a C compiler.