Question

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Double Negation in C++ code

Let’s say:

bool var = !!true;

It will assign “true” to the variable. Seems useless, but I was looking at Visual Studio’s definition of “assert”, and it is:

#define assert(_Expression) (void)( (!!(_Expression)) || (_wassert(_CRT_WIDE(#_Expression), _CRT_WIDE(__FILE__), __LINE__), 0) )

Why does it negate the “_Expression” twice?

I wonder that they want to force the “!” operator to be called (in the case it is overloaded), but that doesn’t seem to be a good reason.

Answer 1

!! guarantees that the result will end up as a 1 or a 0, rather than just the value of _Expression or 0. In C, it’s unlikely to matter, but in C++ I think it turns the result of the expression into a bool type, which might be useful in some cases. If you did have some API that required a literal 1 or 0 be passed to it, using !! would be a way to make it happen.