CODEWITHSUNDEEP
c++c++11
mike
mike

Reputation: 1720

What is the difference between std::initializer_list<int>({1,2,3}) and {1,2,3}?

I have following template function:

template<typename T> void foo2(T t) {}

I know that I cannot call it using:

foo2({1,2,3});

because initializer list is a non deduced context for a template argument. I have to use:

foo2<std::initializer_list<int>>({1,2,3});

but I can also use:

foo2(std::initializer_list<int>({1,2,3}));

which makes me wonder what is the difference between: {1,2,3} and std::initializer_list<int>({1,2,3})?

Upvotes: 5

Views: 405

Answers (1)

Rakete1111
Rakete1111

Reputation: 49028

A braced-init list isn't an expression, and thus doesn't have a type. When you call

foo2({1,2,3});

the compiler doesn't know what type {1,2,3} represents in your mind, and so it doesn't compile.

foo2<std::initializer_list<int>>({1,2,3});

compiles because here the compiler doesn't have to deduce the type, you have specified it, it is std::initializer_list<int>. So it can initialize t with {1,2,3}.

The third call also compiles because the compiler can deduce the type. std::initializer_list<int>({1,2,3}) is obviously a std::initializer_list<int>, and so it can initialize t with the passed prvalue.

Upvotes: 8

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