CODEWITHSUNDEEP
c++c++11templateslanguage-lawyertype-deduction
pooya13
pooya13

Reputation: 2671

C++ - Ambiguous Overloaded Template Resolution/Deduction

Running the following code snippet, I do not receive any errors and I get my expected result. However, since the second template instantiation is ambiguous (both type specifiers are references), I am worried that this might not be defined behavior.
Is this behavior (that the compiler instantiates the most specific overloaded template) guaranteed?

#include <algorithm>
#include <iostream>
#include <vector>

template<typename T>
void Print(const T& x)
{
    std::cout << x << std::endl;
}

template<typename T>
void Print(const std::vector<T>& x)
{
    for(auto it = x.begin(); it != x.end(); ++it)
    {
        std::cout << *it << " ";
    }
    std::cout << std::endl;
}

int main(int argc, char const *argv[])
{
    std::vector<int> v = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    Print(5); // 5
    Print(v); // 0 1 2 3 4 5 6 7 8 9

    return 0;
}

I am not sure where to look so a good reference is also greatly appreciated.

Upvotes: 0

Views: 91

Answers (1)

R Sahu
R Sahu

Reputation: 206567

const std::vector<T>& x is a more specialized type than const T& x. A more specialized type is considered a better match for overload resolution. Hence, your code behaves like it should.

Upvotes: 4

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