题目

We all make mistakes, let's move on.
(don't take this too seriously, no fancy hacking skill is required at all)

This task is based on real event
Thanks to dhmonkey

hint : operator priority

ssh mistake@pwnable.kr -p2222 (pw:guest)

题解

#include <stdio.h>
#include <fcntl.h>

#define PW_LEN 10
#define XORKEY 1

void xor(char* s, int len){
	int i;
	for(i=0; i<len; i++){
		s[i] ^= XORKEY;
	}
}

int main(int argc, char* argv[]){
	
	int fd;
	if(fd=open("/home/mistake/password",O_RDONLY,0400) < 0){
		printf("can't open password %d\n", fd);
		return 0;
	}

	printf("do not bruteforce...\n");
	sleep(time(0)%20);

	char pw_buf[PW_LEN+1];
	int len;
	if(!(len=read(fd,pw_buf,PW_LEN) > 0)){
		printf("read error\n");
		close(fd);
		return 0;		
	}

	char pw_buf2[PW_LEN+1];
	printf("input password : ");
	scanf("%10s", pw_buf2);

	// xor your input
	xor(pw_buf2, 10);

	if(!strncmp(pw_buf, pw_buf2, PW_LEN)){
		printf("Password OK\n");
		system("/bin/cat flag\n");
	}
	else{
		printf("Wrong Password\n");
	}

	close(fd);
	return 0;
}

这题漏洞在条件判断的优先级

以上代码等价于if(fd = 1 < 0), 但是<优先级大于=, 所以if里相当于做了一次赋值, 赋值表达式真值为1, 所以这个条件恒成立, 然后结果是fd == 0, 所以效果就是会从stdin读入两次数据, 那么下面的检测就可以通过了

A的binary为01000001, @的binary为01000000
所以输入AAAAAAAAAA, 和@@@@@@@@@@即可通过检测读取flag