Why Not operator for negative number gives this output

Question

Can somebody please explain output of below code-

int a =-3;
printf("%d",!a);

The output is 0.
I cannot understand why i get output as 0.

If you don't understand that, you have to study what the `!` operator does, simple as that. And also study other utterly fundamental things like how an `if` statement works. This site isn't an interactive beginner tutorial replacing studies. — Lundin, Jun 10 '20 at 06:51

score 0 · Answer 1 · answered Jun 10 '20 at 06:54

Quoting C11, chapter 6.5.3.3

The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int. The expression !E is equivalent to (0==E).

In your case, it can be seen as

 printf("%d", (a == 0));   // where a is -3

which evaluates to a falsy result, thereby printing 0 as the result.

Why Not operator for negative number gives this output

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