Why Template Function call is ambiguous?

Question

#include <iostream>
using namespace std;
template <typename T>
T max(T x, T y)
{
    return (x > y) ? x : y;
}
int main()
{
    cout << max(3, 7) << std::endl;
    cout << max(3.0, 7.0) << std::endl;
    cout << max(3, 7.0) << std::endl;
    return 0;
}

I'm Expecting max's Instance here

cout << max(3, 7) << std::endl; // max (int, int)
cout << max(3.0, 7.0) << std::endl; // max (double, double)
cout << max(3, 7.0) << std::endl; // max(int, double)

Then what is the problem ? Why I'm getting

11 25 [Error] call of overloaded 'max(double, double)' is ambiguous

You should try again without `using namespace std` to remove a source of confusion. — juanchopanza, Sep 08 '15 at 16:45
@juanchopanza `11 28 Desktop\Objects.cpp [Error] no matching function for call to 'max(int, double)' ` — , Sep 08 '15 at 16:52
@LetDoit That is not the same error as "call of overloaded max is ambiguous"! — Barry, Sep 08 '15 at 16:54
Wouldn't the `typename T` be common to the first and the second argument, thereby making the definition `max(int, double)` not possible? — John Bupit, Sep 08 '15 at 16:54
@LetDoit That is because you two need two different template parameters as you are using two different types. — NathanOliver, Sep 08 '15 at 16:54
@NathanOliver here is the code http://codepaste.net/rkh612 why I'm not getting correct output? — , Sep 08 '15 at 17:23
@LetDoit `cout` is truncating the decimal part as it adds nothing. see: http://coliru.stacked-crooked.com/a/961547c050a01ccc — NathanOliver, Sep 08 '15 at 17:27
@LetDoit There is a thing out there called google that has a lot of information. If you googled `cout traling zeros` you would get http://stackoverflow.com/questions/17341870/correct-use-of-stdcout-precision-not-printing-trailing-zeros — NathanOliver, Sep 08 '15 at 17:32

score 5 · Accepted Answer · edited May 23 '17 at 12:14

If you view the compile error in full, you'd see why. Here's what gcc 5.2 gives me:

main.cpp: In function 'int main()':
main.cpp:10:21: error: call of overloaded 'max(int, int)' is ambiguous
     cout << max(3, 7) << std::endl;
                     ^
main.cpp:4:3: note: candidate: T max(T, T) [with T = int]
 T max(T x, T y)
   ^
In file included from /usr/local/include/c++/5.2.0/bits/char_traits.h:39:0,
                 from /usr/local/include/c++/5.2.0/ios:40,
                 from /usr/local/include/c++/5.2.0/ostream:38,
                 from /usr/local/include/c++/5.2.0/iostream:39,
                 from main.cpp:1:
/usr/local/include/c++/5.2.0/bits/stl_algobase.h:219:5: note: candidate: constexpr const _Tp& std::max(const _Tp&, const _Tp&) [with _Tp = int]
     max(const _Tp& __a, const _Tp& __b)
     ^

Basically, there are two max functions - yours and std::max, which is included by some chain of other #includes from <iostream>. The latter is found by lookup because of your

using namespace std;

Effectively, we have:

template <typename T> T max(T, T);                      // yours
template <typename T> T const& max(T const&, T const&); // std

Neither is better than the other, hence ambiguous. This is a great reason to avoid using namespace std. Or a great reason to not reinvent the wheel when it comes to standard library functions - just use std::max. Or both.

On the other hand, this one

max(3, 7.0)

will fail regardless as a template deduction failure. It would deduce T as int for the first argument, and T as double for the second - but there can only be one T! You would have to explicitly call either max<int>(3, 7.0) or max<double>(3, 7.0) to get around the deduction failure, depending on which of the two arguments you want to cast.

which is best compiler? I heard allot about GCC how to download and use it? — , Sep 08 '15 at 16:53
@LetDoit There is no best compiler. I just presented sample output to indicate that compiler output is valuable. If you want to use gcc, you can just google gcc. — Barry, Sep 08 '15 at 16:57
@LetDoit Why do you think you're getting the incorrect output? That prints `7` three times. — Barry, Sep 08 '15 at 17:32

R Sahu · Answer 2 · 2015-09-08T16:58:59.817

The line

using namespace std;

makes things complicated, for sure. However, even after you remove that line, the problem continues to exist.

The call max(3, 7.0) can be resolved to max<int> or max<double>. To resolve to max<int> a double has to be converted to an int. To resolve to max<double>, an int has to be converted to a double. Since both require a conversion and one conversion cannot be given higher priority than the other, the compiler is not able to resolve which one to use.

You'll have to be explicit which version of max you want to use.

max<int>(3, 7.0) will convert 7.0, a double, to an int.

max<double>(3, 7.0) will convert 3, an int, to a double.

score 0 · Answer 3 · answered Oct 15 '19 at 17:35

0

Max is a library function.just change the max identifier for your function to max1 or any other name that is not a function defined any header files. That should solve your problem.

answered Oct 15 '19 at 17:35

Loveforcode

1

Please, try to format your answer in such a way that it clearly distinguishes code examples from prose. You can enclose code using backticks (`). – WTEDST Oct 15 '19 at 17:54

Why Template Function call is ambiguous?

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