Self modifying code always segmentation faults on

2019-01-08 23:47发布

问题:

i found an article about self modifying code and tried to do some examples, but i get always segmentation faults. As fas as i can understand, there is a violation in memory permissions. The code segment is (r)ead/e(x)ecute and so the attempt of writting results to this fault. Is there a way to test the program either by changing the memory permissions at runtime or before? I'm using linux and the example is written in GAS assembly.

.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:

so after modification suggested by osgx here is a working code.(Actually if you assemble&link&run it crashes but if you watch using gdb it does modifies its code!)

.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 start of memory region to
                                        # change its permissions (smc-enabled)
        andl $0xFFFFF000, %esi          # align to start of a pagesize
        pushl   $7                      # permissions==r|w|x
        pushl   $4096                   # page size
        pushl   %esi                    # computed start address
        call    mprotect

        call    changer                 # function that does smc
        mov     $string, %edx
label:
        push    string_end-string       # this code will be overridden
        push    $buf                    # and never be executed!
        push    $string
        call    memcpy
changer:
        mov     $offset_to_write, %esi  # simple copy bytes algorithm
        mov     $label, %edi
        mov     $0xb, %ecx
loop1:
        lodsb
        stosb
        loop    loop1
        ret
offset_to_write:                        # these instructions will be
        push    $0                      # executed eventually
        call    exit
end:

回答1:

You should to change memory access permissions in runtime.

#include <sys/mman.h>

void *addr  = get_address_of_instruction_pointer();
int  length = 4096;   /* size of a page */

if (mprotect(addr, length, PROT_READ | PROT_WRITE | PROT_EXEC) == 0) {
    /* current code page is now writable and code from it is allowed for execution */
}


回答2:

Modern CPUs have a feature called DEP which prevents execution of code on the stack. Previously, this was possible; now, it is not. The binary is loaded into read-only memory by default.

With that out of the way, you can use the mprotect system call to mark your binary's location in memory as executable - SO LONG AS YOUR CODE IS NOT DEP-PROTECTED. So don't try to put code and the stack and then jump into it.



回答3:

You can also disable write-protection for the entire program by passing the switch -N to the linker. If you are invoking the linker from gcc, pass Wl,-N. If you invoke ld directly, pass -N.