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Segmentation fault in C

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# 1  
Old 03-30-2011
Error Segmentation fault in C
i have this code
Code:
int already_there(char *client_names[], char *username) {
    int i;
    for(i = 0; i<NUM; i++) {
        printf("HERE\n");
        if (strcmp(client_names[i], username)==0) return(1);
    }
    return(0);
}

and i get a segmentation fault, whats wrong here?
# 2  
Old 03-30-2011
Well, assuming:
  1. NUM is the length of client_names
  2. all the client names entries are not null, pointers to valid addresses,
  3. every client name is null terminated,
  4. so is username (a valid pointer to a null terminated string)
then, nothing. Maybe test client_names[i] for null. Incidentally, argv is such an array of pointers to char arrays, and while main() provides argc, it is really derived by an extra argv entry at the end, a null pointer, as described in man 2 exec(). Sometimes, the null term is more reliable, like if NUM changes, as there is no Num needed. Stick in some fprintf(stderr,"%d: . . . ", __LINE__, . . .); lines and log what works, unbuffered.
Last edited by DGPickett; 03-30-2011 at 02:19 PM..
# 3  
Old 03-30-2011
nvm i fixed it all i had to do is

if (client_names[i]== username) return(1);
# 4  
Old 03-30-2011
Humm, you are not actually comparing strings here, all you are doing is comparing pointers. Not very robust.
# 5  
Old 03-30-2011
Quote:
Originally Posted by omega666
nvm i fixed it all i had to do is

if (client_names[i]== username) return(1);
Strings don't work that way. It's not crashing anymore, but it's not actually doing what you wanted either. It's just comparing the pointer, not the contents. Try this:

Code:
int main(void)
{
    char *str1="aaaa";
    char str2[]="aaaa";
    char *str3=str1;
    if(str1 == str2) printf("str1 == str2\n");
    if(str1 == str3) printf("str1 == str3\n");
}

It won't print the first one because str1 and str2 aren't equal. They point to different areas of memory. It will print the second one because str1 and str3 do point to the same area of memory.

As is often the case it's not something wrong with your function as much as something wrong with the data you fed into it. Something's wrong with the variables you passed into the function. Since you, as usual, didn't post the complete code, I'd need psychic powers to help you.

Please post your complete code!
Last edited by Corona688; 03-30-2011 at 03:11 PM..
# 6  
Old 03-30-2011
Quote:
Originally Posted by DGPickett
Well, assuming:
  1. NUM is the length of client_names
  2. all the client names entries are not null, pointers to valid addresses,
  3. every client name is null terminated,
  4. so is username (a valid pointer to a null terminated string)
then, nothing. Maybe test client_names[i] for null. Incidentally, argv is such an array of pointers to char arrays, and while main() provides argc, it is really derived by an extra argv entry at the end, a null pointer, as described in man 2 exec(). Sometimes, the null term is more reliable, like if NUM changes, as there is no Num needed. Stick in some fprintf(stderr,"%d: . . . ", __LINE__, . . .); lines and log what works, unbuffered.
i defined client_names like
char *client_names[NUM];
and only one element was defined
how do i set the ones i am not using to not null ?
# 7  
Old 03-30-2011
Do you not understand the meaning of "complete code"? I hope that's enough. Smilie

To set things to NULL, set them equal to NULL.
Code:
// Starting at one because zero has something in it
for(n=1; n<NUM; n++) { client_names[n]=NULL; }

...but that's not enough to stop crashes. NULLs aren't a warning for the computer -- they're a warning for you. The program must check if things are NULL before you use them.

Code:
int already_there(char *client_names[], char *username) {
    int i;
    for(i = 0; i<NUM; i++) {
        printf("HERE\n");
        // strcmp on NULL will crash
        if(client_names[i] != NULL)
        {
            if (strcmp(client_names[i], username)==0) return(1);
        }
    }
    return(0);
}
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