NOT(~) vs NEGATION(!)

Question

#include <iostream>

using namespace std;
int main(int argc, char *argv[]) 
{
   int i=-5;
   while(~(i))
   {
      cout<<i;
      ++i;
   }

 }

The output is -5,-4,-3,-2. Shouldn't it print values till -1?Why is it only till -2. And please explain me the difference between 'not' and 'negation' operators.When ever I write a program they were the source for bugs.

while(i)

I know that the loop condition will be true for positive and negative i's except 0.

while(!i) vs while(~i)

For what values of 'i' the above two loops get executed?

Answer 1

When i gets to -1, the value of ~i is ~-1, or 0, so the while loop stops executing. The ! operator works because it does something completely different; it results in 1 for 0 values and 0 for all other values. ~ is a bitwise negation.

A little more in detail:

• ~ takes each bit in a number and toggles it. So, for example, 10010₂ would become 01101₂
• -1 is all ones in binary when a two's complement signed integer.
• ~0b…11111111 is 0.

However:

• !0 is 1, !anythingElse is 0
• -1 is not 0
• !-1 is still 0

And if you actually want to loop including i == -1, just use while (i) instead of while (~i).

Answer 2

You are correct about i == -1 being the exit condition: your loop is equivalent to

int i=-5;
while(i != -1)
{
    cout<<i;
    ++i;
}
// i == -1 immediately after the loop

When written this way, it should be clear why -1 is not printed the value is first printed, and only then incremented, that's why -2 is the last value that you print.

The ! operator, on the other hand, will produce 1 only when it is given a zero. That's why the loop would print -1 when the ! operator is used in the loop condition.