leg
Daddy told me I should study arm.
But I prefer to study my leg!
Download : http://pwnable.kr/bin/leg.c
Download : http://pwnable.kr/bin/leg.asm
ssh leg@pwnable.kr -p2222 (pw:guest)
leg.c如下:
#include <stdio.h> #include <fcntl.h> int key1(){ asm("mov r3, pc\n"); } int key2(){ asm( "push {r6}\n" "add r6, pc, $1\n" "bx r6\n" ".code 16\n" "mov r3, pc\n" "add r3, $0x4\n" "push {r3}\n" "pop {pc}\n" ".code 32\n" "pop {r6}\n" ); } int key3(){ asm("mov r3, lr\n"); } int main(){ int key=0; printf("Daddy has very strong arm! : "); scanf("%d", &key); if( (key1()+key2()+key3()) == key ){ printf("Congratz!\n"); int fd = open("flag", O_RDONLY); char buf[100]; int r = read(fd, buf, 100); write(0, buf, r); } else{ printf("I have strong leg :P\n"); } return 0; }
leg.asm如下:
(gdb) disass main Dump of assembler code for function main: 0x00008d3c <+0>: push {r4, r11, lr} 0x00008d40 <+4>: add r11, sp, #8 0x00008d44 <+8>: sub sp, sp, #12 0x00008d48 <+12>: mov r3, #0 0x00008d4c <+16>: str r3, [r11, #-16] 0x00008d50 <+20>: ldr r0, [pc, #104] ; 0x8dc0 <main+132> 0x00008d54 <+24>: bl 0xfb6c <printf> 0x00008d58 <+28>: sub r3, r11, #16 0x00008d5c <+32>: ldr r0, [pc, #96] ; 0x8dc4 <main+136> 0x00008d60 <+36>: mov r1, r3 0x00008d64 <+40>: bl 0xfbd8 <__isoc99_scanf> 0x00008d68 <+44>: bl 0x8cd4 <key1> 0x00008d6c <+48>: mov r4, r0 0x00008d70 <+52>: bl 0x8cf0 <key2> 0x00008d74 <+56>: mov r3, r0 0x00008d78 <+60>: add r4, r4, r3 0x00008d7c <+64>: bl 0x8d20 <key3> 0x00008d80 <+68>: mov r3, r0 0x00008d84 <+72>: add r2, r4, r3 0x00008d88 <+76>: ldr r3, [r11, #-16] 0x00008d8c <+80>: cmp r2, r3 0x00008d90 <+84>: bne 0x8da8 <main+108> 0x00008d94 <+88>: ldr r0, [pc, #44] ; 0x8dc8 <main+140> 0x00008d98 <+92>: bl 0x1050c <puts> 0x00008d9c <+96>: ldr r0, [pc, #40] ; 0x8dcc <main+144> 0x00008da0 <+100>: bl 0xf89c <system> 0x00008da4 <+104>: b 0x8db0 <main+116> 0x00008da8 <+108>: ldr r0, [pc, #32] ; 0x8dd0 <main+148> 0x00008dac <+112>: bl 0x1050c <puts> 0x00008db0 <+116>: mov r3, #0 0x00008db4 <+120>: mov r0, r3 0x00008db8 <+124>: sub sp, r11, #8 0x00008dbc <+128>: pop {r4, r11, pc} 0x00008dc0 <+132>: andeq r10, r6, r12, lsl #9 0x00008dc4 <+136>: andeq r10, r6, r12, lsr #9 0x00008dc8 <+140>: ; <UNDEFINED> instruction: 0x0006a4b0 0x00008dcc <+144>: ; <UNDEFINED> instruction: 0x0006a4bc 0x00008dd0 <+148>: andeq r10, r6, r4, asr #9 End of assembler dump. (gdb) disass key1 Dump of assembler code for function key1: 0x00008cd4 <+0>: push {r11} ; (str r11, [sp, #-4]!) 0x00008cd8 <+4>: add r11, sp, #0 0x00008cdc <+8>: mov r3, pc 0x00008ce0 <+12>: mov r0, r3 0x00008ce4 <+16>: sub sp, r11, #0 0x00008ce8 <+20>: pop {r11} ; (ldr r11, [sp], #4) 0x00008cec <+24>: bx lr End of assembler dump. (gdb) disass key2 Dump of assembler code for function key2: 0x00008cf0 <+0>: push {r11} ; (str r11, [sp, #-4]!) 0x00008cf4 <+4>: add r11, sp, #0 0x00008cf8 <+8>: push {r6} ; (str r6, [sp, #-4]!) 0x00008cfc <+12>: add r6, pc, #1 0x00008d00 <+16>: bx r6 0x00008d04 <+20>: mov r3, pc 0x00008d06 <+22>: adds r3, #4 0x00008d08 <+24>: push {r3} 0x00008d0a <+26>: pop {pc} 0x00008d0c <+28>: pop {r6} ; (ldr r6, [sp], #4) 0x00008d10 <+32>: mov r0, r3 0x00008d14 <+36>: sub sp, r11, #0 0x00008d18 <+40>: pop {r11} ; (ldr r11, [sp], #4) 0x00008d1c <+44>: bx lr End of assembler dump. (gdb) disass key3 Dump of assembler code for function key3: 0x00008d20 <+0>: push {r11} ; (str r11, [sp, #-4]!) 0x00008d24 <+4>: add r11, sp, #0 0x00008d28 <+8>: mov r3, lr 0x00008d2c <+12>: mov r0, r3 0x00008d30 <+16>: sub sp, r11, #0 0x00008d34 <+20>: pop {r11} ; (ldr r11, [sp], #4) 0x00008d38 <+24>: bx lr End of assembler dump. (gdb)
arm架构,只要输入为key1()+key2()+key3()就能得到flag
先看key1
由key1+12可知返回值r0是r3传递过去的
由key1+8可知r3是pc传递过去的
arm架构的pc和x86不一样,它指向的是当前语句之后两条的语句,因此pc传递过去的是0x8ce4,所以key1()就是0x8ce4
再看key2
由key2+32可知返回值r0是r3传递过去的
由key2+22可知r3被加上了4
由key2+20可知r3也是pc传递过去的
那么pc传递过去的是0x8d08,r3加上4之后传递给r0的是0x8d0c,所以key2()就是0x8d0c
再看key3
由key3+12可知返回值r0是r3传递过去的
由key3+8可知r3是lr传递过去的
lr是当函数发生bl调用时,指向当前语句的下一条语句
因此由main+64可知lr传递过去的是0x8d80,所以key3()就是0x8d80
因此最后的key就是0x8ce4+0x8d0c+0x8d80=108400