Why does this function prologue use several instructions to calculate the esp reduction?

Question

I have looked at a few dumps of assembler code and there is this section (found here and here) in the main function:

:    push   %ebp


        

Answer 1

So you want a full walk-through without doing any research yourself? Sounds legit.

main+9: mov $0x0, %eax

Loads the register eax with hex 0 (=dec 0).

main+14: add $0xf, %eax

Adds hex F (= dec 15) to the zero in eax.

main+17: add $0xf, %eax

Adds hex F (= dec 15) to eax again. These three instructions could have also been done by

movl $0x1e, %eax

but who's counting instructions... Anyway, at this point eax contains hex 1E which is dec 30.

main+20: shr $0x4, %eax

Shifts the contents of eax to the right by four bits.

main+23: shl $0x4, %eax

Shifts eax right back. Why? Because this clears the lowest four bits. Now eax contains hex 10 (= dec 16)

main+26: sub %eax, %esp

Substracts eax from esp (the stack pointer). Since

main+6: and $0xfffffff0, %esp

cleared the lower four bits in esp previously, the new esp will be sixteen byte aligned, as per ABI. Why not simply use esp after main+6? Because on x86, the stack grows downwards from the top of memory. Simply masking off the lower bits of esp risks clobbering local variables. Hence the subtraction to grow the stack down to the sixteen byte boundary.