actually i tried a program like this
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a=8,b=3;
a|=b;
printf("%d",a);
return 0;
}
i dont understand what happens here a|=b;
i have used || this which is used for or but i dont understand wat happens above
The || operator is the "logical or" operator. the | operator is the "bitwise OR" operator. The |= operator combines the bitwise OR operator and the assignment operator. Thus the code above is equivalent to the code below.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a=8,b=3;
int c = (a | b);
printf("%d",c);
return 0;
}
So it boils down to what does a | b do? First, write 8 and 3 in base 2. We have 8 pairs of bits. If either of them are 1, the result is a 1. Otherwise it is a 0.
EDIT -- For some reason, I am posting text and the editor thinks it is malformed code. I can't seem to post my textual write-up of bitwise operation and get the editor to understand that I am not posting code that needs to be parsed. Anyway, google for tutorials of "C Bitwise OR operator".
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