#include <initializer_list>
#include <utility>
void foo(std::initializer_list<std::pair<int,int>>) {}
template <class T> void bar(T) {}
int main() {
foo({{0,1}}); //This works
foo({{0,1},{1,2}}); //This works
bar({{0,1}}); //This warns
bar({{0,1},{1,2}}); //This fails
bar(std::initializer_list<std::pair<int,int>>({{0,1},{1,2}})); //This works
}
This doesn't compile in gcc 4.5.3, it gives a warning for the marked line saying deducing ‘T’ as ‘std::initializer_list<std::initializer_list<int> >’
and an error for the marked line saying no matching function for call to ‘bar(<brace-enclosed initializer list>)’
. Why can gcc deduce the type of the first call to bar but not the second, and is there a way to fix this other than long and ugly casting?
GCC according to C++11 cannot deduce the type for either first two calls to
bar
. It warns because it implements an extension to C++11.The Standard says that when a function argument in a call to a function template is a
{ ... }
and the parameter is notinitializer_list<X>
(optionally a reference parameter), that then the parameter's type cannot be deduced by the{...}
. If the parameter is such ainitializer_list<X>
, then the elements of the initializer list are deduced independently by comparing againstX
, and each of the deductions of the elements have to match.In this example, the first is OK, and the second is OK too because the first element yields type
int
, and the second element compares{2}
againstT
- this deduction cannot yield a constradiction since it doesn't deduce anything, hence eventually the second call takesT
asint
. The third cannot deduceT
by any element, hence is NOT OK. The last call yields contradicting deductions for two elements.One way to make this work is to use such a type as parameter type
I should note that doing
std::initializer_list<U>({...})
is dangerous - better remove those(...)
around the braces. In your case it happens to work by accident, but considerThe reason is that
({1, 2, 3})
calls the copy/move constructor ofinitializer_list<int>
passing it a temporaryinitializer_list<int>
associated with the{1, 2, 3}
. That temporary object will then be destroyed and die when the initialization is finished. When that temporary object that is associated with the list dies, the backing-up array holding the data will be destroyed too (if the move is elided, it will live as long as "v"; that's bad, since it would not even behave bad guaranteedly!). By omitting the parens,v
is directly associated with the list, and the backing array data is destroyed only whenv
is destroyed.List initialization relies on knowing what type is being initialized.
{1}
could mean many things. When applied to anint
, it fills it with a 1. When applied to astd::vector<int>
, it means to create a one-elementvector
, with1
in the first element. And so on.When you call a template function who's type is completely unconstrained, then there is no type information for list initialization to work with. And without type information, list initialization cannot work.
For example:
You expect this to be of the type
std::initializer_list<std::pair<int,int>>
. But how could the compiler know that?bar
's first parameter is an unconstrained template; it can literally be any type. How could the compiler possibly guess that you meant this specific type?Quite simply, it can't. Compilers are good, but they're not clairvoyant. List initialization can only work in the presence of type information, and unconstrained templates remove all of that.
By all rights, that line should have failed to compile, according to C++11. It cannot deduce what type you intended the
{...}
to be, so it should have failed. That looks like a GCC bug or something.