解释汇编指令

Question

我想翻译如下：

Action: 
pushl %ebp 
movl %esp, %eax 
subl $0x32, %esp 
movl $0x0, -0x8(%eax) 
movl $0x0, -0x4(%eax) 
movl -0x4(%eax), %eax 
cmpl $0x32(%eax), %ebp 
movl -0x4(%ebp), %eax 
sall $0x2, %ebp 
addl 0x8(%ebp), %ebp 
movl (%ebp), %ebp 
addl %ebp, -0x8(%eax) 
addl $0x1, -0x4(%eax) 

什么是翻译这个代码的最佳方式？

Answer 1

对于原题：

mov  -0x4(%ebp), %eax  # eax = z; 
mov  0xc(%ebp), %edx  # edx = y; 
mov  (%edx, %eax, 4), %eax # eax = *(edx + eax + 4) 
add  $0x3, %eax    # eax += 3 
movb $0x41, (%eax)   # *eax = 'A' 

干运行报表给：

y[z + 4][3] = 'A';