What does this code do in assembly language? Please help.

Question

movl %eax, %edx
shrl $31, %edx
addl %edx, %eax
sarl %eax

The program was written in C and complied on a 64-bit Intel Pentium CPU.

Answer 1

movl %eax, %edx
Copies eax to edx.

shrl $31, %edx
Shifts the value in edx to the right by 31 bits and store in edx.
Since edx is a 32 bit register this effectively reads the sign bit.

addl %edx, %eax
Adds edx to eax.
Since the previous operation stored the sign bit in edx, this adds 1 to the original value if eax is negative, otherwise it does nothing.

sarl %eax
Shift eax to the right by 1 and preserves the sign bit. Basically a division by 2.

So for example if the input( eax ) is -5, the result is:
Copy eax(-5) to edx.
Get the sign bit, edx(-5) is negative, so edx is 1.
Add edx(1) to eax(-5), eax is -4.
Divide eax by 2, eax is -2.
Final result: -2.

If the input is 5.
Copy eax(5) to edx.
Get the sign bit, edx(5) is positive, so edx is 0.
Add edx(0) to eax(5), eax is 5. ( do nothing )
Divide eax by 2, eax is 2.
Final result: 2.

Answer 2

Dennis thank you so much.. Now I understand what it means. I truly appreciate your help