段错误（核心哑）x86 汇编 AT&T 语法

Question

我正在尝试编写一个包含 3 个数组的代码，return 最后一个数组中的最大数量（是的，类似于 ProgramminGroundUp 书中的内容）但我希望函数根据数组大小退出不是当它到达元素零时但是代码给了我

Segment fault (Core Dumped)

我使用 'as' 汇编程序'和 gnu 加载程序 'ld' 这是完整的代码

.section .data
first_data_items:
.long 48,65,49,25,36
second_data_items:
.long 123,15,48,67,25,69
third_data_items:
.long 102,120,156,32,14,78,100
.section .text
.globl _start
_start:
pushl $first_data_items
pushl 
call max  
addl , %esp
pushl $second_data_items
pushl 
call max
addl , %esp 
pushl $third_data_items
pushl 
call max
addl , %esp
movl %eax, %ebx
movl , %eax
int [=11=]x80

.type max, @function
max:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %ecx   #ecx will be the size of the array
movl 12(%ebp), %ebx  #ebx will be the base pointer
movl [=11=], %edi         #edi will be the index
movl 0(%ebx, %edi, 4), %eax   #eax will hold the maximum number
start_loop:
cmpl [=11=], %ecx
je end_loop
incl %edi
movl 0(%ebx, %edi,4), %esi     #esi will hold the current element
cmpl %eax, %esi
jle start_loop
movl %esi, %eax
decl %ecx
jmp start_loop 
end_loop:
movl %ebp, %esp
popl %ebp
ret

Answer 1

我在比较语句之后移动了 decl %ecx（感谢 Michael Petch），使 ecx 在每个循环中递减 1 所以代码将是

.section .data
first_data_items:
.long 48,65,49,25,36
second_data_items:
.long 123,15,48,67,25,69
third_data_items:
.long 102,120,156,32,14,170,100
.section .text
.globl _start
_start:
pushl $first_data_items
pushl 
call max
addl , %esp
pushl $second_data_items
pushl 
call max
addl , %esp
pushl $third_data_items
pushl 
call max
addl , %esp
movl %eax, %ebx
movl , %eax
int [=10=]x80
.type max, @function
max:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %ecx   #ecx will be the size of the array
movl 12(%ebp), %ebx  #ebx will be the base pointer
movl [=10=], %edi         #edi will be the index
movl 0(%ebx, %edi, 4), %eax   #eax will hold the maximum number
start_loop:
cmpl [=10=], %ecx
je end_loop
decl %ecx
incl %edi
movl 0(%ebx, %edi,4), %esi         #esi will hold the current element
cmpl %eax, %esi
jle start_loop
movl %esi, %eax
jmp start_loop
end_loop:
movl %ebp, %esp
popl %ebp
ret

Answer 2

关于我对 lodsd 用法的评论，max 函数示例：

.type max, @function
max:
    pushl   %ebp
    movl    %esp, %ebp
    movl    8(%ebp), %ecx       # ecx will be the size of the array
    movl    12(%ebp), %esi      # esi will be the base pointer
    popl    %ebp                # stack_frame usage complete, restore ebp
    lea     (%esi,%ecx,4), %ecx # ecx = (base pointer + 4*size) (end() ptr)
    mov     [=10=]x80000000,%edi    # current max = INT_MIN
start_loop:
    cmp     %ecx, %esi
    jae     end_loop            # end() ptr reached (esi >= ecx)
    lodsl                       # eax = [esi+=4]
    cmp     %edi, %eax          # check if it is new max
    cmovg   %eax, %edi          # update max as needed (eax > edi)
    jmp     start_loop          # go through whole array
end_loop:
    mov     %edi, %eax          # eax = current_max
    ret

（CMOVcc 需要 i686+ 目标架构）

段错误（核心哑）x86 汇编 AT&T 语法

Segment fault (core dumbed) x86 Assembly AT&T syntax

x86

assembly

gnu