Why write `!!x`, when `x` will do? [duplicate]

Question

I often see experienced programmers write !!x, even though the expected expression is a Boolean (i.e., zero or not zero) and not an integer.

For example, a line from boost:

BOOST_ASSERT(!!p); // where `p` is a pointer

What's the point of !!p when just p will do?

What I understand by a Boolean parameter is an expression converted to a value of integral type, and the value is compared against zero, explicitly or implicitly (with if or its ternary operator equivalent).

Thus, anything that takes a Boolean and expects only 0 or 1 is wrongly implemented, if my understanding of Boolean is correct.

For clarification: it's obvious that ! converts to bool; the question is explicitly asking for why.

Answer 1

By default, BOOST_ASSERT(expr) expands to assert(expr). That C++ assert function takes a scalar argument, not bool. Thus, your statement, "the expected expression is a boolean," is not correct.

Classes in C++ can implement operator bool(); the basic_stream classes are examples of such: assert(std::cin) will not work. The !! operator forces the use of std::cin::operator bool(), and bool after !! will be evaluated to int 0 or 1. This is shorter than assert(bool(std::cin)).

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If I were a boost author, I would expand BOOST_ASSERT(expr) to assert(!!(expr)) - as is done for ( BOOST_LIKELY(!!(expr)) ? ((void)0) ).