相关疑难解决方法(0)

我发现了一篇关于自修改代码的文章,并试图做一些例子,但我总是得到分段错误.就像我能理解的那样,内存权限存在违规行为.代码段是(r)ead/e(x)ecute,因此写入的尝试导致此错误.有没有办法通过在运行时或之前更改内存权限来测试程序？我正在使用linux,这个例子是用GAS汇编编写的.

.extern memcpy
.section .data
string:
        .asciz  "whatever"
string_end:
.section .bss
        .lcomm buf, string_end-string
.section .text
.globl main
main:
        call changer
        mov $string, %edx
label:
        push string_end-string
        push $buf
        push $string
        call memcpy
changer:
        mov $offset_to_write, %esi
        mov $label, %edi
        mov $0xb, %ecx
loop1:
        lodsb
        stosb
        loop loop1
        ret
offset_to_write:
        push 0
        call exit
end:

所以在osgx建议的修改后,这是一个工作代码.(实际上,如果你组装并链接并运行它崩溃,但如果你看着使用gdb,它确实修改了它的代码!)

.extern memcpy
.section .data
string:
        .asciz  "Giorgos"
string_end:
.section .bss
        .lcomm buf, string_end-string
.section .text
.globl main
main:
        lea (main), %esi                # get the …

static void swapAddr(int *numOne, int *numTwo)
{
    int *tmp;

    tmp = numOne;
    numOne = numTwo;
    numTwo = tmp;
}

int main(void)
{
    int a = 15;
    int b = 10;

    printf("a is: %d\n", a);
    printf("Address of a: %p\n", &a);
    printf("b is: %d\n", b);
    printf("Address of b: %p\n", &b);

    swapAddr(&a, &b);
    printf("\n");

    printf("a is: %d\n", a);
    printf("Address of a: %p\n", &a);
    printf("b is: %d\n", b);
    printf("Address of b: %p\n", &b);

    return 0;
}

当我编译并运行这段代码时,输​​出是

a is: 15
Address of a: 0x7fff57f39b98 …