What does this ">>=" operator mean in C?

Question

unsigned long set; /*set is after modified*/ set >>= 1; 

I found this in a kernel system call but I don't understand, how does it work?

Answer 1

The expression set >>= 1; means set = set >> 1; that is right shift bits of set by 1 (self assigned form of >> bitwise right shift operator check Bitwise Shift Operators).

Suppose if set is:

BIT NUMBER    31   n=27        m=17                 0               ▼    ▼           ▼                    ▼ set =         0000 1111 1111 1110 0000 0000 0000 0000 

Then after set >> = 1; variable set becomes:

BIT NUMBER    31   n=26        m=16                 0               ▼     ▼           ▼                   ▼ set =         0000 0111 1111 1111 0000 0000 0000 0000 

Notice the bits number shifted.

Note a interesting point: Because set is unsigned long so this >> operation should be logical shift( unsigned shift) a logical shift does not preserve a number's sign bit.

Additionally, because you are shifting all bits to right (towards lower significant number) so one right shift is = divide number by two.

check this code (just to demonstrate last point):

int main(){  unsigned long set = 268304384UL;  set >>= 1;  printf(" set :%lu \n", set);  set = 268304384UL;  set /= 2;  printf(" set :%lu \n", set);  return 1;  } 

And output:

 set :134152192   set :134152192 

(note: its doesn't means >> and / are both same)

Similarly you have operator <<= for left shift, check other available Bitwise operators and Compound assignment operators, also check section: bit expressions and difference between: signed/arithmetic shift and unsigned shift.