A basic assembly question

Aiolia Lee · February 4th, 2010, 02:04 AM

Hello, Sir

I am reading your book and I have a question to ask you.

In chapter 6, the "if statements" topic, you provide a example, the ifthen.c program .After convert the C code to assembly, it looks like as follow:

Code:

main:
        pushl   %ebp
        movl    %esp, %ebp
        andl    $-16, %esp
        subl    $32, %esp
        movl    $100, 24(%esp)
        movl    $20, 28(%esp)
        movl    24(%esp), %eax
        cmpl    28(%esp), %eax
        jle     .L2
        movl    $.LC0, %eax
        movl    24(%esp), %edx
        movl    %edx, 4(%esp)
        movl    %eax, (%esp)
        call    printf
        jmp     .L3
.L2:
        movl    $.LC0, %eax
        movl    28(%esp), %edx
        movl    %edx, 4(%esp)
        movl    %eax, (%esp)
        call    printf
.L3:
        movl    $0, %eax
        leave
        ret

I noticed that there are two assembly statements,
11 andl $-16, %esp
12 subl $32, %esp
in my opinion, these two statements could be instead by this statement,
subl %48, %esp.
I don't know why the gcc compiler use two statements rather than one. BTW, even the latest gcc compiler with O2 option, these two statements still exist.

Hope you reply.
Thank you very much!

bert · May 18th, 2010, 01:00 AM

I also get confused

pengzhixi · December 23rd, 2012, 08:43 PM

11 andl $-16, %esp
12 subl $32, %esp

it is used to align the memory of stack.

bcbarnes · December 25th, 2012, 07:29 AM

To elaborate on what pengzhixi said:

Line 11 is used to mask off the lower 4 bits of ESP. $-16 in hex is 0xFFFFFFF0. This insures that the ESP register is an even multiple of 16, which can allow the processor to optimize the memory accesses.

Line 12 subtracts and additional 32 from the stack, presumably to make room for and locals that have been defined. Note that 32 is also a multiple of 16, and so the stack remains 16 byte aligned.