when using templates C++ "const" mystery overloaded function call - c++

Question

I have a question with the code below. When I compile it it complains with this error below. Now my question is if I remove the "const" in the declaration and change it to only :
template<class T> T max( T& left, T& right);
it becomes okay and the compiles/executes correctly. How come the compiler thinks that the call is ambiguous? Shouldn't the forward declaration follow the implementation?
==== start of error message====
max.cpp:10:34: error: call of overloaded ‘max(int, int)’ is ambiguous
max.cpp:10:34: note: candidates are:
max.cpp:5:21: note: T max(const T&, const T&) [with T = int]
/usr/include/c++/4.6/bits/stl_algobase.h:210:5: note: const _Tp& std::max(const _Tp&, const _Tp&) [with _Tp = int]
max.cpp:11:44: error: call of overloaded ‘max(double, double)’ is ambiguous
max.cpp:11:44: note: candidates are:
max.cpp:5:21: note: T max(const T&, const T&) [with T = double]
=======end of error message===
=============code starts here======
#include<iostream>
using namespace std;
template<class T> T max(const T& left, const T& right);
int main(void)
{
cout<<"max int:"<< max(1,4)<<endl;
cout<<"max double:"<< max(5.02,1.002)<<endl;
}
template<class T>
T max(const T& left, const T& right)
{
return left > right? left:right;
}

Answer 1

You have to do:
cout<<"max int:"<< ::max(1,4)<<endl;
cout<<"max double:"<< ::max(5.02,1.002)<<endl;
Since there is another max defined in std namespace. You have to inform the compiler to use the max defined in global namespace, which is the one you defined. See a live example here: Function template Demo
If you look at the std::max defined in std namespace:
template <class T> const T& max (const T& a, const T& b);
which returns const reference and takes const reference to parameters, that's why when you remove const, it will compile. Hope that this helps.

Answer 2

The std namespace has a max already built in, so you can either not have the line using namespace std; or put the max function in your own namespace.
namespace myNamespace {
//your max function
}

Answer 3

We have few functions already defined in std namespace like max(), swap() qsort(). So when you define your own function
max(const T& left,const T& right)
it is same as one in std namespace hence the error. So you can do following to make it work
call your function using scope resolution operator ::max(x,y)
remove const qualifier from your function

Answer 4

As others have written the compiler has found a max function additionally in namespace std. Now it tries to get the best version for you via function overloading. As both signatures are eaqual it can not choose a best version and complains. Now if you remove const the compiler can not bind a integer (or double) literal to a non const refference. Now it has only one choice and takes std::max functionality. You can see it with the following example:
#include <iostream>
namespace ns1 {
template<class T>
void doit(T const&) {
std::cout << "ns1::doit" << std::endl;
}
}
template<class T>
void doit(T&) {
std::cout << "::doit" << std::endl;
}
int main() {
using namespace ns1;
int i=1;
int const j=1;
doit(1);
doit(i);
doit(i+1-1);
doit(j);
return 0;
}
As everithing is an int the compiler has to choose a non const reference only if it has a non const variable (i). In other cases it choose the cont reference version from namespace ns1. It could even bind temporaries to const references (i+1-1).The same happens with your max and std::max.
Edit:
I forgott to say if you remove the const in your declaration the compiler choose std::max which is of course correct but not yours.