我正在尝试将此Memmove C代码进行汇编,而没有得到预期的结果。
我在xubuntu上使用x86-64程序集,调试了2个小时后,我看不到哪里错了。
C记忆代码:
#include <stdio.h>
extern void * memmove(void *dest, void *src, size_t n);
int main () {
char str1[] = "Geeks"; // Array of size 6
char str2[] = "Quiz"; // Array of size 5
puts("str1 before memmove ");
puts(str1);
/* Copies contents of str2 to sr1 */
memmove(str1, str2, sizeof(str2));
puts("\nstr1 after memmove ");
puts(str1);
return 0;
}
/*
void * memmove(void *dest, void *src, size_t n) {
char *d = (char *) dest;
char *s = (char *) src;
if(s == d)
return dest;
if(s < d) {
//copy from back
s=s+n-1;
d=d+n-1;
while(n--) {
*d-- = *s--;
}
}
else {
//copy from front
while(n--)
*d++ = *s++;
}
return dest;
} */
组装代码:
.globl memmove
# RDI = dest
# RSI = src
# RDX = n
# R8 = d
# R9 = s
memmove:
mov %rdi, %r8 # d = dest
mov %rsi, %r9 # s = src
jmp if_equal
if_equal:
cmp %r8, %r9 # s == d
jz retDest
ja else # s > d
jb if_s_minor # s < d
if_s_minor:
add %rdx, %r9 # s = s + n
sub $1, %r9 # s = s - 1
add %rdx, %r8 # d = d + n
sub $1, %r8 # d = d - 1
jmp while1
while1:
cmp $0, %rdx # n > 0 ?
jna retDest # if n <= 0 go to retDest
sub $1, %rdx # n--
movb (%rsi), %cl # *dst-- = *src--
movb %cl, (%rdi)
leaq -1(%r8), %r8 # *d--
leaq -1(%r9), %r9 # *s--
jmp while1
else:
jmp while2
while2:
cmp $0, %rdx # n > 0 ?
jna retDest
sub $1, %rdx # n--
movb (%rsi), %cl # *dst = *src
movb %cl, (%rdi)
leaq 1(%r8), %r8 # *d++
leaq 1(%r9), %r9 # *s++
jmp while2
retDest:
mov %rdi, %rax
ret
.end
本来应该在第二张照片上显示“ Quiz”,但它显示了这一点:
str1在记忆之前 极客
记忆后的str1 Qeeks
答案 0 :(得分:1)
movb (%rsi), %cl # *dst = *src movb %cl, (%rdi) leaq 1(%r8), %r8 # *d++ leaq 1(%r9), %r9 # *s++ jmp while2
这里的问题是您的代码更改了%r8
和%r9
寄存器,但是实际移动停留中使用的%rdi
和%rsi
寄存器相同,从而一次又一次地重复复制相同的字节!
while1 代码中当然也存在同样的问题。