Any way to print out template typename value?

[RRick]

I have a simple template method that looks like

    template < typename T>
    void WhatEver( void)
    {
         T t;
         // How to print with cout the T value (not the t value)?
    }

I thought maybe RTTI would work, but that deals with classes with virtual methods. Since T is most likely going to be an int or double, is there any built in C++ way to print out the value of T.

[Banfa]

I assume you mean to actually output the typename used by the class in which case the answer is basically no.

[mac11]

It might not be exactly what you want, but gcc provides __PRETTY_FUNCTI ON__ which ends up printing the type of T for you:

#include <iostream>
#include <string>
using namespace std;

template <typename T>
void WhatEver(void)
{
    T t;
    cout << __PRETTY_FUNCTION__ << endl;
}

int main()
{
    WhatEver<int>();
    WhatEver<long>();
    WhatEver<string>();

    return 0;
}

output:

void WhatEver() [with T = int]
void WhatEver() [with T = long int]
void WhatEver() [with T = std::basic_stri ng<char, std::char_trait s<char>, std::allocator< char> >]

If you aren't using gcc maybe you can look under it's hood to find out how __PRETTY_FUNCTI ON__ works?

[weaknessforcats]

This is C++. Use the typeid keyword to see the type of the variable at run time:

template < typename T> 
    void WhatEver( void) 
    { 
         T t; 
         // How to print with cout the T value (not the t value)? 
         cout << typeid(t).name() << endl;    //Your answer
    } 
int main()
{
    WhatEver<int>();
    WhatEver<double>();
}

Research C++ RTTI for further info.

[RRick]

This works fine, but the output is not very recognizable. With this format, typeid gives you a name of i for int, j for unsigned int, c for char, and (oh yes) h for unsigned char.

GNU g++ uses the Itaniium C++ ABI (http://www.codesourcery.com/public/cxx-abi/abi.html). Go past the mangled name section for more info about the encodings.

It appears to be a format designed to print out any variable defined in the know universe. Whew!

[weaknessforcats]

The name() method is implementation dependent. The idea is that youonly use it for debugging. Otherwise, you are supposed to compare the type_info object returned by typeid with one created from a known type. The removes the dependence on the implementation.

if ( typeid(int) == typeid(myVar) )
{
     //etc...
}

BTW: Visual Studio typeid name() method returns int for int unsigned char for unsigned char, etc.