对time_t和int64的重载函数进行模棱两可的调用

Question

I am getting following error

1>------ Build started: Project: test123, Configuration: Debug Win32 ------
1>  test.cpp
1>e:\avinash\test123\test.cpp(25): error C2668: 'XYZ::createKey' : ambiguous call to overloaded function
1>          e:\avinash\test123\test.cpp(13): could be 'void *XYZ::createKey(const int64_t)'
1>          e:\avinash\test123\test.cpp(7): or       'void *XYZ::createKey(const time_t &)'
1>          while trying to match the argument list '(long)'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Following is the source code, how do i solve this

#include <WinSock2.h>
typedef signed __int64   int64;
typedef int64            int64_t;

namespace XYZ 
{
    inline void* createKey( const time_t& value ) {
        return NULL;
    }
    inline void* createValue( const time_t& value ) {
        return NULL;
    }
    inline void* createKey(const int64_t value) {                     
        return NULL;
    }                                                                     
    inline void* createValue(const int64_t value) {                      
        return NULL;
    }  
}



int main( int argc, char** argv) 
{
    XYZ::createKey(10L);
    return 0;
}

Answer 1

time_t is __int64 on your platform. It might be long or int on other platforms. There is no way to define separate overloads for time_t and the primitive integer type it aliases, because it is not a separate type .

The overloads above are pointless. The compiler won't change them based on whether the argument is declared time_t or __int64 but based on whether the argument is a const reference or not (if I read the specification correctly the reference is a worse match for anything except exactly the same reference type than the non-reference, but I am not sure about it by far).