void(os << args). What does void mean in this context?-CodePudding

I was reading about folding expressions and found an example of what was used before folding expressions:

template <class... Ts>
void print_all(std::ostream& os, Ts const&... args) {
    using expander = int[];
    (void)expander{0,
        (void(os << args), 0)...
    };
}

The problem is the void(os << args) bit. What does void mean in this context? I've already tried to search for this, but it is pretty generic.

Thanks for your time.

CodePudding user response：

It casts the result of (os << args) to void. That's it.

This style is used to prevent the very rare overload of ,(comma) operator since in theory, << can be overloaded for user-defined args argument and return something that has overloaded the comma operator for X, 0 expression. That might break the initialization trick. This way, the code is safe and comma is always used as the ordinary sequencing operator.