What does this expression means: bool(__args...)

user_7588455 0
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I just stumbled around in the STL for C++ and found a quite surprising exception in the vector header.
Cleaned up from some unnecessary junk it boils down to this:
(This snippets could be found in the vector class within the STL vector header)

template<typename... _Args>
reference emplace_back(_Args&&... __args) {
    push_back(bool(__args...));
    return back();
}

void push_back(bool __x) {
    if (this->_M_impl._M_finish._M_p != this->_M_impl._M_end_addr())
        *this->_M_impl._M_finish++ = __x;
    else
         _M_insert_aux(end(), __x);
}

As already outlined in the title, I'm quite surprised by the bool(__args...) expression which makes no sense for me at all.
I hope somebody could bring light into the darkness and help me out.

Edit: Answered, I was a bit lazy

Solution found, I was a bit lazy.
I used my IDEs method resolver to jump to the declaration.
But sadly it directed me to the stl_bvector.h header and the given method above.
Since it seems that this header only handles the implementation for std::vector<bool> the code makes much more sense.

stl c++
1 Answers

I think you're checking the implementation of std::vector<bool>.

bool(__args...) is constructing a temporary bool from the parameter pack __args... (which is expanded into comma-separated expressions), then the temporary bool is passed to the push_back(bool).

user_33097900
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