Question

Quick question on left shifts in assembly using the “sall” instruction.

From what I understand, “sall rightop, leftop” would translate to “leftop = leftop << rightop”, so taking an integer and shifting the bits 4 spaces to the left would result in a multiplication by 2^4.

But what happens when the integer is unsigned, 32-bits, and is something like:

1111 1111 1111 1111 1111 0000 0010 0010

Would a left shift in this case become 1111 1111 1111 1111 0000 0010 0010 0000 ?

Obviously this is not a multiplication by 2^4.

Thanks!!

Answer 1

It is a multiplication by 2^4, modulo 2^32:

n = (n * 2^4) % (2 ^ 32)

You can detect the bits that got “shifted out” by performing a shift left followed by masking, in this case

dropped = (n >> (32-4)) & (1<<4-1)