VC6和VS2008之间的模板功能行为

Question

I have simple code about template function for Visual C++ 6.0 and Visual Studio 2008.

#include <stdio.h>
#include <vector>

template<typename T>
void function(const std::vector<T> &vec)
{
    printf("vector version\n");
}

template<typename T>
void function(T val)
{
    printf("value version\n");
}

int main()
{
    std::vector<int> vec;
    function(vec);

    return 0;
}

I tried for each environment, and finally get
at VC6, function of value version, and
at VS2008, function of vector version.

I have 2 questions.

1. I have recognized priority of overloaded function call as following,
   a) specialized function (without implicit type convert)
   b) template function (without implicit type convert)
   c) specialized function, with implicit type convert
   d) template function, with implicit type convert

   with this rule, the above results seems that
   at VC6, b) is accepted (with <T> = std::vector<int>)
   at VS2008, b) is ignored(?) and d) is accepted(?) (with <T> = int)

   This means that VC6 is valid and VS2008 is wrong.
   Is not my guess correct?

2. Although, I wish vector version is called for both VC6 and VS2008.
   Can I do it?

Regards.

Answer 1

Actually VC6 is wrong; MS had limited support for the C++99 standard (which is when templates were standardized) in VC6 and had better support in VS2005 and up.

Calling function(vec) calls

template<typename T>
void function(const std::template vector<T>& vec)

with T as an int type because the template was deduced from the vector template type (the same as calling function<int>(vec) ). If you called function(&vec) then the value function will be called since you're passing in a reference, which gets deduced to function<std::vector<int>>(vec) .

If you want it to always call the proper function, then you'll need to be explicit, so you'd need to call it as such:

function< std::vector<int> >(vec)

Which will deduce to the vector version. Note the space between the > , this is to avoid the compiler thinking you meant the stream operator >> .

Hope that helps.