我很难将 C 代码转换为程序集。这是我需要转换的代码：
#include <stdio.h>
define N 50 int x[N], y[N], z[2 * N];
void convolve(int[], int[], int[], int);
int main(void)
{
int i, n;
printf("Enter vector size (<=%d): ", N);
scanf("%d", &n);
printf("Enter first vector (%d elements):\n", n);
for (i = 0; i < n; i++)
**scanf("%d", &x[i]);
**printf("Enter second vector (%d elements):\n", n);
for (i = 0; i < n; i++)
scanf("%d", &y[i]);
convolve(x, y, z, n);
printf("Convolution:\n");
for (i = 0; i < ((n + n) - 1); i++)
printf("%d ", z[i]);
printf("\n");
return 0;
}
void convolve(int x[], int y[], int z[], int n)
{
int i, j;
for (i = 0; i < ((n + n) - 1); i++)
z[i] = 0;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
z[i + j] += x[i] * y[j];
return;
}
我被困在这一行：
scanf("%d", &x[i]);
如何插入数组？
这是我到目前为止所做的：
.data
.align 4
state: .long 0
.bss
N = 50
int x[N], y[N], z[2*N]
.data
.equ N, 50
.comm i,4,4 #int b
.comm n,4,4 #int n
.comm j,4,4 #int j
.comm x,N*4,4 #int x[N] where N is 50
.comm y,N*4,4 #int x[N] where N is 50
.comm z,N*8,4 #int x[N] where N is 100
.section .rodata #to format strings
fmt0: .string "Enter vector size (<=%d): "
fmt1: .string "%d"
fmt2: .string "Enter first element (%d elements):\n"
fmt3: .string "Enter second element (%d elements):\n"
fmt4: .string "Convolution:\n"
fmt5: .string "\n"
fmt6: .string .%d .
.text
.globl main
main:
pushl %ebp #prolog
movl %esp, %ebp
pushl %esi #save callee-save registers %esi, %edi, and %ebx onto stack
pushl %edi # where %esi at -4(%ebp),%edi at -8(%ebp), and %ebx at -12(%ebp)
pushl %ebx
pushl %eax #for array where %eax at -16(%ebp)------------------------------
/* Allocate space for i and n on the stack */
subl $8, %esp
/* i is at address -20(%ebp) */
/* n is at address -24(%ebp) */
pushl $fmt0 #push fmt0
call printf #printf("Enter vector size (<=%d): ")
addl $4, %esp #deallocate parm to printf
leal -24(%ebp), %ebx #%ebx = address of n
pushl %ebx #push address of n
pushl $fmt1 #push fmt1 "%d"
call scanf #scanf ("%d", &n)
addl $8, %esp #dealoccate parms for scanf
pushl $fmt2 #push fmt2
call printf #printf("Enter first element (%d elements):\n")
addl $4, %esp #deallocate parm to printf
movl $0, -20(%ebp) #i=0
movl -20(%ebp), %edi #%edi=i
movl -24(%ebp), %esi #esi=n
cmpl %esi, %edi #compare i:n
jg for_done #jump to for_done if i>n
for_loop:
pushl %edi #push i
pushl %esi #push n
pushl %eax #push array
pushl $fmt1 #push fmt1 ("%d")
call scanf #scanf("%d", n)
addl $8, %esp #dealocate parms to scanf
movl (address of x,%edi,4), %eax------------------------------------------------------
incl %edi #%edi++ (i++)
movl %edi,-20(%ebp) #i=%edi
compl %esi, %edi #compare i:n
jle for_loop #jump to for_loop if i<n
for_done:
addl $8, %esp #deallocate local vars from stack
popl %ebx #restore %ebx
popl %edi #restore %edi
popl %esi #restore %esi
/*next loop for second vector*/
pushl %esi #save callee-save registers %esi, %edi, and %ebx onto stack
pushl %edi # where %esi at -4(%ebp),%edi at -8(%ebp), and %ebx at -12(%ebp)
pushl %ebx
pushl $fmt3 #push fmt3
call printf #printf("Enter second element (%d elements):\n")
addl $4, %esp #deallocate parm to printf
movl $0, -20(%ebp) #i=0
movl -20(%ebp), %edi #%edi=i
movl -24(%ebp), %esi #esi=n
cmpl %esi, %edi #compare i:n
jg for_done #jump to for_done if i>n
for_loop:
pushl %edi #push i
pushl %esi #push n
pushl %eax #push array
pushl $fmt1 #push fmt1 ("%d")
call scanf #scanf("%d", n)
addl $8, %esp #dealocate parms to scanf
movl (address of y,%edi,4), %eax------------------------------------------------------
incl %edi #%edi++ (i++)
movl %edi,-20(%ebp) #i=%edi
compl %esi, %edi #compare i:n
jle for_loop #jump to for_loop if i<n
for_done:
addl $8, %esp #deallocate local vars from stack
popl %ebx #restore %ebx
popl %edi #restore %edi
popl %esi #restore %esi
leave #epilog
ret
convolve:
pushl %ebp #prolog
movl %esp, %ebp
pushl %esi #save callee-save registers %esi, %edi, and %ebx onto stack
pushl %edi # where %esi at -4(%ebp),%edi at -8(%ebp), and %ebx at -12(%ebp)
pushl %ebx
/* Allocate space for x, y, z, n, i, and j on the stack */
subl $24, %esp
/* x is at address 4(%ebp) */
/* y is at address 8(%ebp) */
/* z is at address 12(%ebp) */
/* n is at address 16(%ebp) */
/* i is at address -16(%ebp) */
/* n is at address -20(%ebp) */
movl $0, -16(%ebp) #i=0
movl -16(%ebp), %edi #%edi=i
movl -20(%ebp), %esi #esi=n
addl %esi, %esi #2 times n
subl $1, %esi #2n - 1
cmpl %esi, %edi #compare i:n
jg for_done #jump to for_done if i>n
答案 0 :(得分：1)
我不禁注意到你已经定义了两个不同的for_loop和for_done;你可能想以某种方式区分它们。
根据其他一些建议，也许像这样的东西，例如，可以工作：
shuffle_batch...
for_loop1:
pushl %edi #push i
pushl %esi #push n
leal x(,%edi,4), %eax
pushl %eax #push array