在C中调用crypt函数时出现分段错误

Question

I pass the argument Arturo $1$salt$ and I get a Segmentation fault (core dumped) error.

#define _XOPEN_SOURCE
#include <stdio.h>
#include <unistd.h>
#include <string.h>

int main (int argc, char* argv[])
{
    if ( argc != 3) {
        printf ("Usage: ./crypt key salt\n");
        return 1;
    }

    printf ("%s\n", crypt (argv[1], argv[2]));

    return 0;
}

Answer 1

The answer turned out to be simple - bash variable expansion. the $ character is reserved in the shell, to mark the beginning of a variable

so when you run

./program Arturo $1$salt$

the argv[2] after variable expansion will be

"$"

which is not a valid salt after the glibc specification (which expects $id$salt$ ). The call to crypt with that seed will return NULL and set errno to EINVAL , because the seed is invalid, and the call to printf chokes on the NULL and segfaults, which is the behaviour you are seeing.

if you were to execute your program as follows, disabling variable expansion in the shell:

./program Arturo '$1$salt$'

the output would be

$1$salt$y5SOwLketmwNfSvW0yAoz/

as expected.

Answer 2

Assuming:

typedef char *string;

From crypt POSIX documentation :

The salt argument shall be a string of at least two bytes in length not including the null character chosen from the set:

abcdefghijklmnopqrstu vwxyz ABCDEFGHIJKLMNOPQRSTU VWXYZ 0 1 2 3 4 5 6 7 8 9 . /

but you pass $1$salt$ .

So you have to check the return value of crypt and print only if is ! = NULL ! = NULL .

Now, if you want to use crypt function from glibc, you have to include crypt.h header in your program. See example in glibc documentation .