#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?
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)
.
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.
'~' is the operator that : ~x = -x-1
and when i = -1, then ~i = 0.
if you wonder the value of ~i, you can just print them out:
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int i=-5;
for (int i = -5; i <= 3; i++)
{
cout<<i<<" "<<(~i)<<endl;
}
}
and then you will find:
-5 4
-4 3
-3 2
-2 1
-1 0
0 -1
1 -2
2 -3
3 -4