Pwnable-collision

The same connection ssh, enter the password to view the file

Col.c look at the source code

#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;
}

For check_password hashcode function's return value is equal to res There In Flag, i.e. hashcode == 0x21DD09EC == res, check_password look function, turn into strong int, and five times the accumulated output res, while a second main function of the following if a constraint length of 20, an int is 4 bytes, divided into five groups exactly 20 bytes

 

We 0x21DD09EC to decimal 568 134 124, for convenience of calculation is again divided by adding 5 gave 113,626,825, before deduction of plus one is 1136268254, i.e. (113,626,825 113,626,8254 + = 568 134 124 * 4)

 

Then they turn into hexadecimal 0x6c5cec9 * 4 + 0x6c5cec8 = 0x21dd09ec

 

To know the value of the input can be input here for little endian input python

./col $(python -c 'print "\xc9\xce\xc5\x06"*4+"\xc8\xce\xc5\x06"')

 

 daddy! I just managed to create a hash collision :)