Bit manipulation in c

Question

int samp=0;

for(i=0;i<=31;i++)
{
  samp=samp|1<<i;
}
printf("\ %d\n",samp);

output:

-1

Why is it giving -1 if I loop till i<=31 (setting all 32 bits to 1)? When I loop only i<31 it is giving 2147483647. Why is it so?

Answer 1

This is happening because the first bit is the sign bit.

When the sign bit is 1 , the number is negative, and 11111111 11111111 11111111 11111111 happens to be the 32-bit representation of the number -1 .

You may want to check out Two's Complement .

Answer 2

In your printf statement you are using %d , which prints a signed integer . The output is correct due to the sign bit being set.

Change the format string to %u and it will display the unsigned integer value. No more sign bit and the value you are looking for.

You should be using an unsigned int anyway for samp .

Answer 3

由于现代计算机中的带符号整数表示为两个补码： http : //en.wikipedia.org/wiki/Two%27s_complement