Arch: amd64-64-little<br>
RELRO: Partial RELRO<br>
Stack: No canary found<br>
NX: NX enabled<br>
PIE: PIE enabled<br>
以下是main函数初始化完后的栈构造
除了Canary之外的保护都开了，由于开启了PIE无法直接获悉具体指令地址，所以无法直接构建ROP<br>
system调用位于main函数中低2字节为0x0AD2处
位于RBP-0x18处是main函数地址<br>
位于RSP+0x10处是VDSO基址
或在GDB中 vmmap<br>
0xffffffffff600000 0xffffffffff601000 --xp 1000 0 [vsyscall]
或cat /proc/1/maps | grep vsyscal<br>
ffffffffff600000-ffffffffff601000 --xp 00000000 00:00 0 [vsyscall]
对于这个实现方法，成功在本地打通并且getShell，但是在远程却并不适用，题目名叫做Ret2VDSO，直接使用Vsyscall我也觉得应该不是出题人本意，但是我并没有完全理解VDSO和该程序的利用，也没有找到相关类似题目的WP，根据我对vsyscall和vdso的理解推测或许是靶机上压根就没有vsyscall，所以对于栈溢出的利用仅到此为止；
我曾尝试爆破VDSO基址，并加上0x5FC作为偏移来作为RetGadGet，但是这个方法不论是在远程还是本地都没成功，原因是在x64下，该地址有3个字节也既12位是随机的，不同于x32下仅有1字节，爆破概率是1/256；在本程序中爆破概率低达1/16777216，在绞尽脑汁想不出如何通过栈溢出来打通该程序时，确实抱着侥幸尝试过爆破VDSO基址，但不论本地还是远程最终都没能打通，即使打通了本地，也许我找的VDSO文件的ret偏移0x5FC在远程上也并不适用，所以我最终放弃了
根据分析可知以下三点
由于程序中的seed根据time()来生成，所以这是可以被预估的，只需要提前预估未来的某一个符合rand结果小于9位数用户可输入的时间戳并且等时间到时发送内容即可
我对该机制的理解还是过于浅显，以至于无法用出题人希望的方式解出这道题，目 前这题作为遗留问题，在我理解深入后再回过头来思考这题的解法
find / -name '*vdso*.so*'
int getint()
{
//buf大小为16字节
char buf[16]; // [rsp+0h] [rbp-10h]
//仅允许用户输入8个字节
read(0, buf, 8uLL);
return atoi(buf);//将输入转为数字并返回
}
int menu()
{
puts("------welcome------");
puts("1.get gift");
puts("2.overflow");
puts("3.exit");
puts("[+]give me your choice:");
return getint();
}
int __cdecl main(int argc, const char **argv, const char **envp)
{
unsigned int seed; // eax
int inputNum; // ebx
char buf[56]; // [rsp+0h] [rbp-50h] BYREF
int choice; // [rsp+38h] [rbp-18h]
int counter; // [rsp+3Ch] [rbp-14h]
setvbuf(_bss_start, 0LL, 2, 0LL);
setvbuf(stdin, 0LL, 1, 0LL);
counter = 2;
do
{
if ( !counter )//如果counter为0则结束循环
break;
choice = menu();//获取用户输入
if ( choice == 1 )
{
--counter;
puts("input num:");
seed = time(0LL);//将当前时间戳作为seed【不安全的引用】
srand(seed);
inputNum = getint();//获取用户输入【注意是先获取seed后才等待输入】
if ( inputNum == rand() )//对比输入和rand结果，如果一致则直接getshell
system("/bin/sh");
}
if ( choice == 2 )
{
--counter;
puts("hello from ctfhub");
read(0, buf, 0xD0uLL);//【栈溢出】
}
}
while ( choice != 3 );//即使输入3也会再跑一次循环【逻辑错误】
return 0;
}
int getint()
{
//buf大小为16字节
char buf[16]; // [rsp+0h] [rbp-10h]
//仅允许用户输入8个字节
read(0, buf, 8uLL);
return atoi(buf);//将输入转为数字并返回
}
int menu()
{
puts("------welcome------");
puts("1.get gift");
puts("2.overflow");
puts("3.exit");
puts("[+]give me your choice:");
return getint();
}
int __cdecl main(int argc, const char **argv, const char **envp)
{
unsigned int seed; // eax
int inputNum; // ebx
char buf[56]; // [rsp+0h] [rbp-50h] BYREF
int choice; // [rsp+38h] [rbp-18h]
int counter; // [rsp+3Ch] [rbp-14h]
setvbuf(_bss_start, 0LL, 2, 0LL);
setvbuf(stdin, 0LL, 1, 0LL);
counter = 2;
do
{
if ( !counter )//如果counter为0则结束循环
break;
choice = menu();//获取用户输入
if ( choice == 1 )
{
--counter;
puts("input num:");
seed = time(0LL);//将当前时间戳作为seed【不安全的引用】
srand(seed);
inputNum = getint();//获取用户输入【注意是先获取seed后才等待输入】
if ( inputNum == rand() )//对比输入和rand结果，如果一致则直接getshell
system("/bin/sh");
}
if ( choice == 2 )
{
--counter;
puts("hello from ctfhub");
read(0, buf, 0xD0uLL);//【栈溢出】
}
}
while ( choice != 3 );//即使输入3也会再跑一次循环【逻辑错误】
return 0;
}
.text:0000000000000A38 ; int __cdecl main(int argc, const char **argv, const char **envp)
.text:0000000000000A38 public main
.text:0000000000000A38 main proc near ; DATA XREF: _start+1D↑o
.text:0000000000000A38
.text:0000000000000A38 buf= byte ptr -50h
.text:0000000000000A38 choice= dword ptr -18h
.text:0000000000000A38 counter= dword ptr -14h
.text:0000000000000A38
.text:0000000000000A38 ; __unwind {
.text:0000000000000A38 push rbp
.text:0000000000000A39 mov rbp, rsp
.text:0000000000000A3C push rbx
.text:0000000000000A3D sub rsp, 48h
.text:0000000000000A41 ; 8: setvbuf(_bss_start, 0LL, 2, 0LL);
.text:0000000000000A41 mov rax, cs:__bss_start
.text:0000000000000A48 mov ecx, 0 ; n
.text:0000000000000A4D mov edx, 2 ; modes
.text:0000000000000A52 mov esi, 0 ; buf
.text:0000000000000A57 mov rdi, rax ; stream
.text:0000000000000A5A call _setvbuf
.text:0000000000000A5A
.text:0000000000000A5F ; 9: setvbuf(stdin, 0LL, 1, 0LL);
.text:0000000000000A5F mov rax, cs:stdin@@GLIBC_2_2_5
.text:0000000000000A66 mov ecx, 0 ; n
.text:0000000000000A6B mov edx, 1 ; modes
.text:0000000000000A70 mov esi, 0 ; buf
.text:0000000000000A75 mov rdi, rax ; stream
.text:0000000000000A78 call _setvbuf
.text:0000000000000A78
.text:0000000000000A7D ; 10: v8 = 2;
.text:0000000000000A7D mov [rbp+counter], 2
.text:0000000000000A84 jmp loc_B10
.text:0000000000000A84
.text:0000000000000A89 ; ---------------------------------------------------------------------------
.text:0000000000000A89 ; 15: choice = getInput();
.text:0000000000000A89
.text:0000000000000A89 loc_A89: ; CODE XREF: main+DC↓j
.text:0000000000000A89 mov eax, 0
.text:0000000000000A8E call printMenu ; 打印菜单并获取用户输入数字
.text:0000000000000A8E ; 1：getGift，判断用户输入和随机数是否相同
.text:0000000000000A8E ; 2：overflow，栈溢出
.text:0000000000000A8E ; 3：结束
.text:0000000000000A8E
.text:0000000000000A93 mov [rbp+choice], eax
.text:0000000000000A96 ; 16: if ( choice == 1 )
.text:0000000000000A96 cmp [rbp+choice], 1
.text:0000000000000A9A jnz short loc_ADE
.text:0000000000000A9A
.text:0000000000000A9C ; 18: --v8;
.text:0000000000000A9C sub [rbp+counter], 1
; ↓↓↓↓↓↓↓↓↓↓用户输入为1，getGift↓↓↓↓↓↓↓↓↓↓
.text:0000000000000AA0 ; 19: puts("input num:");
.text:0000000000000AA0 lea rdi, aInputNum ; "input num:"
.text:0000000000000AA7 call _puts
.text:0000000000000AA7
.text:0000000000000AAC ; 20: seed = time(0LL);
.text:0000000000000AAC mov edi, 0 ; timer
.text:0000000000000AB1 call _time
.text:0000000000000AB1
.text:0000000000000AB6 ; 21: srand(seed);
.text:0000000000000AB6 mov edi, eax ; seed
.text:0000000000000AB8 call _srand
.text:0000000000000AB8
.text:0000000000000ABD ; 22: inputNum = getInputNumber();
.text:0000000000000ABD mov eax, 0
.text:0000000000000AC2 call getInputNumber
.text:0000000000000AC2
.text:0000000000000AC7 mov ebx, eax
.text:0000000000000AC9 ; 23: if ( inputNum == rand() )
.text:0000000000000AC9 call _rand
.text:0000000000000AC9
.text:0000000000000ACE cmp ebx, eax
.text:0000000000000AD0 jnz short loc_ADE
.text:0000000000000AD0
.text:0000000000000AD2 ;system("/bin/sh");getShell代码位于mainAD2处
.text:0000000000000AD2 lea rdi, command ; "/bin/sh"
.text:0000000000000AD9 call _system
; ↑↑↑↑↑↑↑↑↑↑↑用户输入为1:getGift↑↑↑↑↑↑↑↑↑↑↑↑↑
.text:0000000000000AD9
.text:0000000000000ADE ; 26: if ( choice == 2 )
.text:0000000000000ADE
.text:0000000000000ADE loc_ADE: ; CODE XREF: main+62↑j
.text:0000000000000ADE ; main+98↑j
.text:0000000000000ADE cmp [rbp+choice], 2
.text:0000000000000AE2 jnz short loc_B0A
; ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓用户输入为2：overflow↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
.text:0000000000000AE2
.text:0000000000000AE4 ; 28: --v8;
.text:0000000000000AE4 sub [rbp+counter], 1
.text:0000000000000AE8 ; 29: puts("hello from ctfhub");
.text:0000000000000AE80 lea rdi, aHelloFromCtfhu ; "hello from ctfhub"
.text:0000000000000AEF call _puts
.text:0000000000000AEF
.text:0000000000000AF4 ; 30: read(0, buf, 0xD0uLL);
.text:0000000000000AF4 lea rax, [rbp+buf]
.text:0000000000000AF8 mov edx, 0D0h ; nbytes
.text:0000000000000AFD mov rsi, rax ; buf
.text:0000000000000B00 mov edi, 0 ; fd
.text:0000000000000B05 call _read
; ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑用户输入为2：overflow↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑
.text:0000000000000B05
.text:0000000000000B0A ; 33: while ( choice != 3 );
.text:0000000000000B0A
.text:0000000000000B0A loc_B0A: ; CODE XREF: main+AA↑j
.text:0000000000000B0A cmp [rbp+choice], 3
.text:0000000000000B0E jz short loc_B1C
.text:0000000000000B0E
.text:0000000000000B10 ; 13: if ( !v8 )
.text:0000000000000B10
.text:0000000000000B10 loc_B10: ; CODE XREF: main+4C↑j
.text:0000000000000B10 cmp [rbp+counter], 0
.text:0000000000000B14 ; 14: break;
.text:0000000000000B14 jnz loc_A89
.text:0000000000000B14
.text:0000000000000B1A jmp short loc_B1D
.text:0000000000000B1A
.text:0000000000000B1C ; ---------------------------------------------------------------------------
.text:0000000000000B1C
.text:0000000000000B1C loc_B1C: ; CODE XREF: main+D6↑j
.text:0000000000000B1C nop
.text:0000000000000B1C
.text:0000000000000B1D ; 34: return 0;
.text:0000000000000B1D
.text:0000000000000B1D loc_B1D: ; CODE XREF: main+E2↑j
.text:0000000000000B1D mov eax, 0
.text:0000000000000B22 add rsp, 48h
.text:0000000000000B26 pop rbx
.text:0000000000000B27 pop rbp
.text:0000000000000B28 retn
.text:0000000000000B28 ; } // starts at A38
.text:0000000000000B28
.text:0000000000000B28 main endp
.text:0000000000000A38 ; int __cdecl main(int argc, const char **argv, const char **envp)
.text:0000000000000A38 public main
.text:0000000000000A38 main proc near ; DATA XREF: _start+1D↑o
.text:0000000000000A38
.text:0000000000000A38 buf= byte ptr -50h
.text:0000000000000A38 choice= dword ptr -18h
.text:0000000000000A38 counter= dword ptr -14h
.text:0000000000000A38
.text:0000000000000A38 ; __unwind {
.text:0000000000000A38 push rbp
.text:0000000000000A39 mov rbp, rsp
.text:0000000000000A3C push rbx
.text:0000000000000A3D sub rsp, 48h
.text:0000000000000A41 ; 8: setvbuf(_bss_start, 0LL, 2, 0LL);
.text:0000000000000A41 mov rax, cs:__bss_start
.text:0000000000000A48 mov ecx, 0 ; n
.text:0000000000000A4D mov edx, 2 ; modes
.text:0000000000000A52 mov esi, 0 ; buf
.text:0000000000000A57 mov rdi, rax ; stream
.text:0000000000000A5A call _setvbuf
.text:0000000000000A5A
.text:0000000000000A5F ; 9: setvbuf(stdin, 0LL, 1, 0LL);
.text:0000000000000A5F mov rax, cs:stdin@@GLIBC_2_2_5
.text:0000000000000A66 mov ecx, 0 ; n
.text:0000000000000A6B mov edx, 1 ; modes
.text:0000000000000A70 mov esi, 0 ; buf
.text:0000000000000A75 mov rdi, rax ; stream
.text:0000000000000A78 call _setvbuf
.text:0000000000000A78
.text:0000000000000A7D ; 10: v8 = 2;
.text:0000000000000A7D mov [rbp+counter], 2
.text:0000000000000A84 jmp loc_B10
.text:0000000000000A84
.text:0000000000000A89 ; ---------------------------------------------------------------------------
.text:0000000000000A89 ; 15: choice = getInput();
.text:0000000000000A89
.text:0000000000000A89 loc_A89: ; CODE XREF: main+DC↓j
.text:0000000000000A89 mov eax, 0
.text:0000000000000A8E call printMenu ; 打印菜单并获取用户输入数字
.text:0000000000000A8E ; 1：getGift，判断用户输入和随机数是否相同
.text:0000000000000A8E ; 2：overflow，栈溢出
.text:0000000000000A8E ; 3：结束
.text:0000000000000A8E
.text:0000000000000A93 mov [rbp+choice], eax
.text:0000000000000A96 ; 16: if ( choice == 1 )
.text:0000000000000A96 cmp [rbp+choice], 1
.text:0000000000000A9A jnz short loc_ADE
.text:0000000000000A9A
.text:0000000000000A9C ; 18: --v8;
.text:0000000000000A9C sub [rbp+counter], 1
; ↓↓↓↓↓↓↓↓↓↓用户输入为1，getGift↓↓↓↓↓↓↓↓↓↓
.text:0000000000000AA0 ; 19: puts("input num:");
.text:0000000000000AA0 lea rdi, aInputNum ; "input num:"
.text:0000000000000AA7 call _puts
.text:0000000000000AA7
.text:0000000000000AAC ; 20: seed = time(0LL);
.text:0000000000000AAC mov edi, 0 ; timer
.text:0000000000000AB1 call _time
.text:0000000000000AB1
.text:0000000000000AB6 ; 21: srand(seed);
.text:0000000000000AB6 mov edi, eax ; seed
.text:0000000000000AB8 call _srand
.text:0000000000000AB8
.text:0000000000000ABD ; 22: inputNum = getInputNumber();
.text:0000000000000ABD mov eax, 0
.text:0000000000000AC2 call getInputNumber
.text:0000000000000AC2
.text:0000000000000AC7 mov ebx, eax
.text:0000000000000AC9 ; 23: if ( inputNum == rand() )
.text:0000000000000AC9 call _rand
.text:0000000000000AC9
.text:0000000000000ACE cmp ebx, eax
.text:0000000000000AD0 jnz short loc_ADE
.text:0000000000000AD0
.text:0000000000000AD2 ;system("/bin/sh");getShell代码位于mainAD2处
.text:0000000000000AD2 lea rdi, command ; "/bin/sh"
.text:0000000000000AD9 call _system
; ↑↑↑↑↑↑↑↑↑↑↑用户输入为1:getGift↑↑↑↑↑↑↑↑↑↑↑↑↑
.text:0000000000000AD9
.text:0000000000000ADE ; 26: if ( choice == 2 )
.text:0000000000000ADE
.text:0000000000000ADE loc_ADE: ; CODE XREF: main+62↑j
.text:0000000000000ADE ; main+98↑j
.text:0000000000000ADE cmp [rbp+choice], 2
.text:0000000000000AE2 jnz short loc_B0A
; ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓用户输入为2：overflow↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
.text:0000000000000AE2
.text:0000000000000AE4 ; 28: --v8;
.text:0000000000000AE4 sub [rbp+counter], 1
.text:0000000000000AE8 ; 29: puts("hello from ctfhub");
.text:0000000000000AE80 lea rdi, aHelloFromCtfhu ; "hello from ctfhub"
.text:0000000000000AEF call _puts
.text:0000000000000AEF
.text:0000000000000AF4 ; 30: read(0, buf, 0xD0uLL);
.text:0000000000000AF4 lea rax, [rbp+buf]
.text:0000000000000AF8 mov edx, 0D0h ; nbytes
.text:0000000000000AFD mov rsi, rax ; buf
.text:0000000000000B00 mov edi, 0 ; fd
.text:0000000000000B05 call _read
; ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑用户输入为2：overflow↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑
.text:0000000000000B05
.text:0000000000000B0A ; 33: while ( choice != 3 );
[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)
最后于 2023-3-13 09:48
被LeaMov编辑
，原因： 上传样本，修改md格式
