What is the purpose of a double negative in macro definition, like (!!(expr))? [duplicate]

Answer 1

It is a way to cast an expression to bool. In C++ though, operator! can be overloaded. Another way for C/C++:

0 != (expr)

C++ only way:

static_cast<bool>(expr)

[Edit] Thinking more about C++ operator overloading, it make sense to avoid using operators. Libraries like Boost.Spirit and Boost.Lambda use expression templates and lazy evaluation, so that expressions like (expr) || call() may behave not as expected. The most bullet proof version of that macro looks like this:

#define uassert(expr) if(expr) {} else { uasserted(...); }

Here, only conversion of expr to bool is used. The else branch is needed to protect from expressions like uassert(x) else something_else();.

Answer 2

In the example you showed us, the !! is utterly useless. However, in general it's used as a pre-C99 rough equivalent to casting to _Bool, i.e. 0 stays 0 and any other value becomes 1.

Answer 3

It will just make sure that the expr part of the macro is converted to a bool.

Answer 4

It is an idiom sometimes used in C++ for converting to a boolean type.