Bit Setting and Bit Shifting in Ansi C

Question

Can anyone explain this bitwise operation syntax?

#define Bitset(var,bitno) ((var) |=1UL<<(bitno))

I know it sets the bits of var, but I can't understand the syntax.

Answer 1

Let's break it down, piece by piece:

1UL is an unsigned long int with a value of 1 represented at the bit level as:

00000000000000000000000000000001

the << is a "bit shift" operator which will move all the bits in that value above to the left bitno number of times. If it's 1UL<<5, you'll end up with:

00000000000000000000000000100000

Once you have this value, the |= (which is a bitwise OR operation with an assignment) will essentially force the bit of var that's in line with that 1 to be a 1 and wont touch any other bits because (X | 0 = X)

Lets say var is 37 and bitno is 7. Then everything at the bit level will look like this:

00000000000000000000000000100101  // var
00000000000000000000000010000000  // 1UL<<7

00000000000000000000000010100101  // var | (1UL<<7)

Finally, in case it isn't clear, the #define marks Bitset as a function-like macro.