CODEWITHSUNDEEP
c++c++11initializer-list
Person.Junkie
Person.Junkie

Reputation: 1896

convert 'std::initializer_list<int>' to 'int'

Why I can not use variable assigned using "initializer_list" as "normal" variable?

code:

void stovr(int a){}
int main() {

   auto v {5}; // v is std::initializer_list<int>
   stovr(v);  // cannot convert 'std::initializer_list<int>' to 'int'

}

Upvotes: 3

Views: 1499

Answers (2)

Drop
Drop

Reputation: 13013

"A braced initializer has no type! A braced initializer has no type! A braced initializer has no type! No type corresponds to braced initializer"

(C) Scott Meyers.

This is why auto type deduction of brace-initializer with one element fails to follow intuitive rules in C++11 and C++14.

It is considered a defect in standard by many programmers. That is why proposal N3922: New Rules for auto deduction from braced-init-list exists. Notably:

For direct list-initialization:

  • For a braced-init-list with only a single element, auto deduction will deduce from that entry;
  • For a braced-init-list with more than one element, auto deduction will be ill-formed.

Not sure if it is a breaking change and will all C++14 code work after this.

Source: There was a good talk on CppCon about quirks of type deduction in C++11/C++14/C++17 by Scott Meyers: link, including brace-initializer auto type deduction (starting from 29:00). Same thing in more details in his new book (and there was a free sampler with that chapter as well)

Upvotes: 2

user743382
user743382

Reputation:

std::initializer_list<T> cannot convert to T for the same reason that T[] cannot convert to T, and vector<T> cannot convert to T: if you merely know that you have a std::initializer_list<T>, you don't know how many elements it has.

std::initializer_list<int> x = { 1, 2, 3 }; is perfectly valid. If there were an implicit conversion from std::initializer_list<int> to int, which of the values would you expect to see?

Upvotes: 7

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