Double Negation in C++

It's a trick to convert to bool.

It's actually a very useful idiom in some contexts. Take these macros (example from the Linux kernel). For GCC, they're implemented as follows:

#define likely(cond)   (__builtin_expect(!!(cond), 1))
#define unlikely(cond) (__builtin_expect(!!(cond), 0))

Why do they have to do this? GCC's __builtin_expect treats its parameters as long and not bool, so there needs to be some form of conversion. Since they don't know what cond is when they're writing those macros, it is most general to simply use the !! idiom.

They could probably do the same thing by comparing against 0, but in my opinion, it's actually more straightforward to do the double-negation, since that's the closest to a cast-to-bool that C has.

This code can be used in C++ as well... it's a lowest-common-denominator thing. If possible, do what works in both C and C++.

The coders think that it will convert the operand to bool, but because the operands of && are already implicitly converted to bool, it's utterly redundant.

Yes it is correct and no you are not missing something. !! is a conversion to bool. See this question for more discussion.

It's a technique to avoid writing (variable != 0) - i.e. to convert from whatever type it is to a bool.

IMO Code like this has no place in systems that need to be maintained - because it is not immediately readable code (hence the question in the first place).

Code must be legible - otherwise you leave a time debt legacy for the future - as it takes time to understand something that is needlessly convoluted.

It side-steps a compiler warning. Try this:

int _tmain(int argc, _TCHAR* argv[])
{
    int foo = 5;
    bool bar = foo;
    bool baz = !!foo;
    return 0;
}

The 'bar' line generates a "forcing value to bool 'true' or 'false' (performance warning)" on MSVC++, but the 'baz' line sneaks through fine.

Is operator! overloaded?
If not, they're probably doing this to convert the variable to a bool without producing a warning. This is definitely not a standard way of doing things.

As Marcin mentioned, it might well matter if operator overloading is in play. Otherwise, in C/C++ it doesn't matter except if you're doing one of the following things:

• direct comparison to true (or in C something like a TRUE macro), which is almost always a bad idea. For example:

  if (api.lookup("some-string") == true) {...}

• you simply want something converted to a strict 0/1 value. In C++ an assignment to a bool will do this implicitly (for those things that are implicitly convertible to bool). In C or if you're dealing with a non-bool variable, this is an idiom that I've seen, but I prefer the (some_variable != 0) variety myself.

I think in the context of a larger boolean expression it simply clutters things up.

This may be an example of the double-bang trick (see The Safe Bool Idiom for more details). Here I summarize the first page of the article.

In C++ there are a number of ways to provide Boolean tests for classes.

An obvious way is the operator bool conversion operator.

// operator bool version
class Testable {
    bool ok_;
    public:
    explicit Testable(bool b = true) : ok_(b) {}

    operator bool() const { // use bool conversion operator
      return ok_;
    }
};

We can test the class as thus:

Testable test;
if (test) {
    std::cout << "Yes, test is working!\n";
}
else { 
    std::cout << "No, test is not working!\n";
}

However, operator bool is considered unsafe because it allows nonsensical operations such as test << 1; or int i = test.

Using operator! is safer because we avoid implicit conversion or overloading issues.

The implementation is trivial,

bool operator!() const { // use operator!
    return !ok_;
}

The two idiomatic ways to test Testable object are

Testable test;
if (!!test) {
    std::cout << "Yes, test is working!\n";
}
if (!test) {
    std::cout << "No, test is not working!\n";
}

The first version if (!!test) is what some people call the double-bang trick.

It's correct but, in C, pointless here -- 'if' and '&&' would treat the expression the same way without the '!!'.

The reason to do this in C++, I suppose, is that '&&' could be overloaded. But then, so could '!', so it doesn't really guarantee you get a bool, without looking at the code for the types of variable and api.call. Maybe someone with more C++ experience could explain; perhaps it's meant as a defense-in-depth sort of measure, not a guarantee.

Maybe the programmers were thinking something like this...

!!myAnswer is boolean. In context, it should become boolean, but I just love to bang bang things to make sure, because once upon a time there was a mysterious bug that bit me, and bang bang, I killed it.