CODEWITHSUNDEEP
c++c++11
Moustapha Ramadan
Moustapha Ramadan

Reputation: 109

Ambiguous call to overloaded function in variadic template function

I am using a variadic template function where the function parameters isn't the templated types.

I got a compilation error:

Error C2668 '_TailHelper': ambiguous call to overloaded function

Here it is the code snippet.

template <typename HEAD>
void _TailHelper(int) {
    std::cout << typeid(HEAD).name() << std::endl;
}

template <typename HEAD, typename ... TAILS>
void _TailHelper(int x) {
    _TailHelper<HEAD>(x);
    _TailHelper<TAILS...>(x);
}


int main(){
    _TailHelper<int,double>(2);
}

Upvotes: 2

Views: 605

Answers (3)

Jarod42
Jarod42

Reputation: 217245

As alternative to recursion, you might "loop" over your variadic:

In C++17:

template <typename... Ts>
void PrintTypes() {
    ((std::cout << typeid(Ts).name() << std::endl), ...);
}

In previous version, it is less elegant, and you might use some trick as:

template <typename... Ts>
void PrintTypes() {
    const int dummy[] = {0, ((std::cout << typeid(Ts).name() << std::endl), 0)...};
    static_cast<void>(dummy); // Avoid warning for unused variable
}

Upvotes: 0

&#214;&#246; Tiib
&#214;&#246; Tiib

Reputation: 10979

Both overloads match with single template argument, so you have to disable one. For example like that:

#include <iostream>    
#include <typeinfo>

template <typename T>
void TailHelper(int) { 
    std::cout << typeid(T).name() << std::endl;
}

template <typename HEAD, typename ... TAILS>
typename std::enable_if<(sizeof...(TAILS) != 0)>::type
TailHelper(int x) {
    TailHelper<HEAD>(x);        
    TailHelper<TAILS...>(x);
}

int main() {
    TailHelper<int,double>(2);
}

Upvotes: 5

avish
avish

Reputation: 48

The ambiguous call comes from this line:

_TailHelper<HEAD>(x);

this call matches both functions the first one, and the second function which its second parameter can refer to zero or more template parameters.

Upvotes: 0

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