我有以下汇编代码:
间接1.s
.section .data
t1:
.long 5
.section .text
.globl _start
_start:
movl $t1, %ecx #we are passing the address to %ecx
movl $5, %eax #we are passing value 5 to %eax
movl (%ecx), %ebx #Using indirect addressing mode we are getting the value from t1 and passing it to ebx
addl %eax, %ebx # add the values in %eax, %ebx and store it in %ebx
movl $1, %eax # call exit program
int $0x80 # Call Master Bruce Wayne
当上述程序运行时,我按预期得到值 10
[ashok@localhost asm-32]$ as indirect1.s -gstabs+ -o indirect1.o
[ashok@localhost asm-32]$ ld indirect1.o -o indirect1
[ashok@localhost asm-32]$ ./indirect1
[ashok@localhost asm-32]$ echo $?
10
修改了上述程序以消除%ecx寄存器:
间接2.s
.section .data
t1:
.long 5
.section .text
.globl _start
_start:
movl $t1, %ebx # we are passing the address to %ebx
movl $5, %eax # we are passing value 5 to %eax
addl %eax, (%ebx) # add the values in %eax, %ebx and store it in %ebx
movl $1, %eax # call exit program
int $0x80 # Call Master Bruce Wayne
当我运行上面的程序时,我没有得到预期的输出,即 10 我似乎获取存储在 %ebx 中的地址
[ashok@localhost asm-32]$ as indirect2.s -gstabs+ -o indirect2.o
[ashok@localhost asm-32]$ ld indirect2.o -o indirect2
[ashok@localhost asm-32]$ ./indirect2
[ashok@localhost asm-32]$ echo $?
136
我在间接2.s程序中做错了什么。
我认为你想要的是这样的东西:
movl $t1, %ebx # ebx = address of t1
movl $5, %eax # eax = 5
addl (%ebx), %eax # eax += (ebx)
movl %eax, %ebx # exit value
movl $1, %eax # exit()
int $0x80
或者,要使第二个示例正常工作:
.section .data
t1:
.long 5
.section .text
.globl _start
_start:
movl $t1, %ebx # we are passing the address to %ebx
movl $5, %eax # we are passing value 5 to %eax
addl %eax, (%ebx) # add the values in %eax, %ebx and store it in %ebx
movl (%ebx), %ebx # THE FORGOTTEN INSTRUCTION (read result back into %ebx)
movl $1, %eax # call exit program
int $0x80 # Call Master Bruce Wayne
发生的事情是您的 indirect2 的初始版本打印出$t1
的相对地址,即程序退出时%ebx
的地址。