What is the difference between !! and nothing?


This question already has an answer here:


!!a is almost equivalent to a. It converts it to a boolean value.

Usually this does not make a difference, but sometimes it does.

#include <iostream>

int a(int x) {
    return 1;

int a(bool x) {
    return 2;

int main() {
    std::cout << a(0) << std::endl;   //prints 1
    std::cout << a(!!0) << std::endl; //prints 2

    std::cout << -1 << std::endl;   //prints -1
    std::cout << !!-1 << std::endl; //prints 1


In your case, there is no difference, unless there is overloading. (But even if there is overloading, I hope there is no difference.)

(FYI, this is done even more commonly in Javascript because of its types, e.g. false != null but false == !!null. I include this comment because you tagged your question with C, C++, and Java, and Javascript shares similar syntax.)

If operator! is not overloaded, the two statements are equivalent.

Where !! might be useful is if you need to change a zero / non-zero value, or a nullptr / non-null pointer value into a 0/1 value in an integer expression.

For a (dubious) example, the following loop counts the number of non-zero elements in a vector:

for (size_t i = 0; i != v.size(); i++)  
    count += !!v[i];

You will sometimes see !! in bit-level hacks as a result. But in the if statement you show above? Unless operator! is overloaded, that's not a useful operation.

There must be some form of operator overloading else, it means the same.

Take this as an live example:

#include <stdio.h>

int main(void) {
  int a = 5;

  printf("%d\n", a);
  printf("%d\n", !!a); // will print 1

  return 0;

The main difference is a silly-warning from Visual C++.

Secondary (rare) differences include the case where you have an operator!, and the case where the argument is used in a non-boolean context, e.g. arithmetic.

Double ! simplily means a unary NOT and another unary not. See it as (!(!on_debugging))

Yes, you are right, most of the time the results is the same as on_debugging. I think this is for preciseness or strictness to use !! because the ! operator only returns integer 0 or 1, which corresponds to false and true. While the variable can be anything of int and point type.

For Java !!on_debugging is the opposite of !!!on_debugging.

You can negate stuff as often as you want, thus there is no difference between on_debugging and !!on_debugging.

Also see Java Language Specification for this operator:

The type of the operand expression of the unary ! operator must be boolean or Boolean, or a compile-time error occurs. (see JLS 15.15.6).

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