[Pwnable.kr] collision

by yiz96 21. 2 月 2016 pwnable writeup 0

这道题意思大概是告诉你密码是一个20位的字符串，然后每4个字母的内存转换成一个整数（32位），然后五个整数加起来等于0x21DD09EC即可过关。

#include <stdio.h>
#include <string.h>
unsigned long hashcode = 0x21DD09EC;
unsigned long check_password(const char* p){
        int* ip = (int*)p;
        int i;
        int res=0;
        for(i=0; i<5; i++){
                res += ip[i];
        }
        return res;
}

int main(int argc, char* argv[]){
        if(argc<2){
                printf("usage : %s [passcode]\n", argv[0]);
                return 0;
        }
        if(strlen(argv[1]) != 20){
                printf("passcode length should be 20 bytes\n");
                return 0;
        }

        if(hashcode == check_password( argv[1] )){
                system("/bin/cat flag");
                return 0;
        }
        else
                printf("wrong passcode.\n");
        return 0;
}

#include <stdio.h>

#include <string.h>

unsigned long hashcode = 0x21DD09EC;

unsigned long check_password(const char* p){

int* ip = (int*)p;

int i;

int res=0;

for(i=0; i<5; i++){

res += ip[i];

}

return res;

}

int main(int argc, char* argv[]){

if(argc<2){

printf("usage : %s [passcode]\n", argv[0]);

return 0;

}

if(strlen(argv[1]) != 20){

printf("passcode length should be 20 bytes\n");

return 0;

}

if(hashcode == check_password( argv[1] )){

system("/bin/cat flag");

return 0;

}

else

printf("wrong passcode.\n");

return 0;

}

0x21DD09EC = 4*0x06C5CEC8 + 0x06C5CECC输入即可

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