!! c operator, is a two NOT?

I reading this [code](

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), and have this line

switch (!!up + !!left) {

what is `!!` operator ? two logical NOT ?

yes, it's two nots.

`!!a` is `1` if `a` is non-zero and `0` if `a` is `0`

You can think of `!!` as clamping, as it were, to `{0,1}`. I personally find the usage a bad attempt to appear fancy.

You're right. It's two nots. To see why one would do this, try this code:

#include <stdio.h>

int foo(const int a)
{
return !!a;
}

int main()
{
const int b = foo(7);
printf(
"The boolean value is %d, "
"where 1 means true and 0 means false.\n",
b
);
return 0;
}

It outputs `The boolean value is 1, where 1 means true and 0 means false.` If you drop the `!!`, though, it outputs `The boolean value is 7, where 1 means true and 0 means false.`

You can imagine it like this:

!(!(a))

If you do it step by step, this make sense

result = !42; //Result = 0
result = !(!42) //Result = 1 because !0 = 1

This will return `1` with any number (-42, 4.2f, etc.) but only with `0`, this will happens

result = !0; //Result = 1
result = !(!0) //result = 0

`!!` is a more-portable (pre-C99) alternative to `(_Bool)`.