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Why does the number of elements in a initializer list cause an ambiguous call error?

Posted by: admin November 30, 2017 Leave a comment


Why are the first two calls to doSomething OK by the compiler, but using two elements in the list causes an ambiguous call?

#include <vector>
#include <string>

void doSomething(const std::vector<std::string>& data) {}

void doSomething(const std::vector<int>& data) {}

int main(int argc, char *argv[])
    doSomething({"hello"}); // OK
    doSomething({"hello", "stack", "overflow"}); // OK
    doSomething({"hello", "stack"}); // C2668 'doSomething': ambiguous call

    return 0;

What is happening here is that in the two element initializer list both of the string literals can be implicitly converted to const char* since their type is const char[N]. Now std::vector has a constructor that takes two iterators which the pointers qualify for. Because of that the initializer_list constructor of the std::vector<std::string> is conflicting with the iterator range constructor of std::vector<int>.

If we change the code to instead be

doSomething({"hello"s, "stack"s});

Then the elements of the initializer list are now std::strings so there is no ambiguity.


Both the one-argument and three-argument lists can only match std::vector<std::string>‘s std::initializer_list constructor. However, the two-argument list matches one of the constructors from std::vector<int>:

template <class InputIt>
vector(InputIt first, InputIt last, Allocator const &alloc = Allocator());

Indeed, a char const * can be incremented, and dereferenced to get a char that is implicitly convertible to an int.


"hello" and "stack" both decay to const char * which satisfies the InputIterator concept. This allow’s them to match std::vector‘s constructor #4.

If you pass std::string objects the ambiguity is resolved.