How does C++ determine this should be a string_view?

Question

Consider the following code:

#include <optional>
#include <string_view>

int main() {
    std::optional<std::string_view> opt { "abc" };
    std::cout << opt.value();

    return 0;
}

It compiles and when run outputs "abc" to the console. Why does it compile? The first line of code in main should be:

std::optional<std::string_view> { std::string_view { "abc" } };

How does the compiler know to make a call to the string_view constructor with the literal? This seems like it is doing more than just type deduction which I have seen before. Here it seems like the compiler is adding code to my source, i.e. a call to a constructor.

Answer 1

How does the compiler know to make a call to the string_view constructor with the literal?

std::optional has a constructor with the form of

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

This is used to directly initialize the underlying object only if T can be constructed from U, which it can in this case using std::string_view's

constexpr basic_string_view(const CharT* s);

constructor.

This seems like it is doing more than just type deduction which I have seen before

You told the compiler you have a std::optional<std::string_view>, so it doesn't need to determine what type of optional you have. It does determine what U would be in the constructor, and sees that is a type which can be used to construct a std::string_view with, and so it calls the constructor to do just that since it satisfies the requirements.