What does this strange C++ syntax mean?

Question

This syntax is so strange, I don't know how to understand it.

I don’t know what it means, and I tried searching on major search engines, but the keywords were clueless and no results were found.

Please help me, thanks.

auto parser::error(const std::string& msg) -> void
{
    std::cerr << msg << std::endl;
}

It is a [trailing return](https://en.cppreference.com/w/cpp/language/function#Function_declaration). — rawrex, Jul 07 '21 at 02:56
This should be useful: https://stackoverflow.com/q/388242/2079303 — eerorika, Jul 07 '21 at 05:42

score 2 · Answer 1 · answered Jul 07 '21 at 03:44

This is C++ 11, with auto type deduction syntax for function return type. The trailing part -> void is called as trailing return type syntax

I will refer you to Effective Modern C++ by Scott Meyers. This syntax is detailed in Item-3 "Understand decltype" of this book, so just reading the first-3 items will make this syntax clear to you.

This particular case does not seem to be a good use case for this syntax, and it can simply be written as below

void parser::error(const std::string& msg)
{
    std::cerr << msg << std::endl;
}

What does this strange C++ syntax mean?

1 Answers1