I'm learning assembly by comparing a c program to its assembly equivalent.
Here is the code.
.file "ex3.c"
.section .rodata
.LC0:
.string "I am %d years old.\n"
.LC1:
.string "I am %d inches tall.\n"
.text
.globl main
.type main, @function
main:
pushl %ebp //establish stack frame//
movl %esp, %ebp //move esp into ebp, all contents saved down stack//
andl $-16, %esp //16 from esp for local var space//
subl $32, %esp//stack frame reserving - 32 bytes//
movl $10, 24(%esp)
movl $72, 28(%esp)
movl 24(%esp), %eax
movl %eax, 4(%esp)
movl $.LC0, (%esp)
call printf
movl 28(%esp), %eax
movl %eax, 4(%esp)
movl $.LC1, (%esp)
call printf
movl $0, %eax
leave
ret
.size main, .-main
.ident "GCC: (Ubuntu 4.8.2-19ubuntu1) 4.8.2"
.section .note.GNU-stack,"",@progbits
For this line:
movl $10, 24(%esp)
If I understand it correctly it is saying move the value of 10 into the esp register. But what is the 24 doing? I don't think it is moved into esp because a value to be moved is denoted by "$" (i think)
2 Answers
movl $10,24(%esp)
means: move a literal decimal-10 long (4-bytes) into a 4-byte memory location that begins at the address pointed to by (the esp register plus decimal 24)--basically it is a local variable.
In other words movl $10,24(%esp)
means: load 10 into *(esp + 24)
In C that equals to:
*(unsigned long *)(myptr + 24) = 10;
where myptr is taken with value of esp register.
