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std::optional has the following constructor:

template < class U = T > constexpr optional( U&& value );

The question here is: why template parameter U is defaulted to type T? What happens if simply change constructor to following:

template < class U /* = T */> constexpr optional( U&& value );
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1
  • Triple backticks are only for code blocks, not inline code. Just use one for that. Commented Nov 22, 2022 at 0:25

1 Answer 1

Reset to default
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It's so if you give it an initializer list (which doesn't have a type, so can't infer a type for U), it will initialize a T temporary.

For example:

std::optional<std::vector<int>> opt({1, 2, 3}); // No type deduced for `U`, defaults to `std::vector<int>` struct X { int a, b; }; std::optional<X> opt({.a = 1, .b = 2}); // Same here
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