Question

I have created a template as follows

template<typename T>
void f(T const& t) { }

I wanted for this to be callable by containers but also by initializer lists. I thought it would be initializer_list<int>, when called as follows.

f({1, 2, 3});

But GCC behaves as if it’s not Standards compliant

m.cpp: In function 'int main()':
m.cpp:6:25: warning: deducing 'const T' as 'const std::initializer_list<int>'
m.cpp:4:6: warning:   in call to 'void f(const T&) [with T = std::initializer_list<int>]'
m.cpp:6:25: warning:   (you can disable this with -fno-deduce-init-list)

Can anyone explain how I can make this work without warnings? Thanks!

Answer 1

A “thing” like {1,2,3} does not qualify as an expression. It has no type. Therefore, no type deduction is done. But C++0x makes an explicit exception for ‘auto’, so

auto x = {1,2,3};

actually works and decltype(x) will be initializer_list<int>. But this is a special rule that only applies to auto. I guess they wanted to make loops like these

for (int x : {2,3,5,7,11}) {
   ...
}

work since this kind of loop exploits the special rule.

As for solving the problem, you could add an initializer_list<T> overload as a “wrapper”:

template<class T>
inline void outer(initializer_list<T> il) {
   inner(il);
}

I didn’t test this but my current understanding is that it should work.