#include<stdio.h>
int main()
{
int i=255;
i=i<<24;
i=i>>24;
printf(“%d”,i);
return 0;
}
output is : -1
how??
#include<stdio.h>
int main()
{
int i=255;
i=i<<24;
i=i>>24;
printf(“%d”,i);
return 0;
}
output is : -1
how??
u have taken int and it is four byte long take it as char then check the output.
hi,,,
we have i=255;
its bit pattern is
00000000 00000000 00000000 11111111
now applying only i<<24
00000000 00000000 11111111 00000000 (8) shift
00000000 11111111 00000000 00000000 (16) shift
11111111 00000000 00000000 00000000 (24) shift
now i=i<>24
in right shifting (we follow arithmetic shifting) in which if the leftmost bit is 1 then we will fill it with 0 .if leftmost bit is 1 then it is filled with 1;(this is exception in right shift while there is not such exception in left shifting).
on first 8bit right shifting we get
11111111 11111111 00000000 00000000
on 16 bit shifting we get
11111111 11111111 11111111 00000000
on 24 bit shifting
11111111 11111111 1111111 11111111
after shifting this is assigned to i
i=i>>24;
so i becomes
i=11111111 11111111 1111111 11111111
and in decimal form this is (-1).
but if we take unsigned int then it is giving 255….
why??