null terminate an array of strings

Question

I am trying to figure out how to get my array of strings from get_arguments to NULL terminate, or if that isn't the issue to function in my execv call.

char ** get_arguments(const char * string) {
    char * copy = strdup(string);
    char * remove_newline = "";
    for(;;) {
        remove_newline = strpbrk(copy, "\n\t");
        if (remove_newline) {
            strcpy(remove_newline, "");
        }
        else {
            break;
        }
    }   
    char (* temp)[16] = (char *) malloc(256 * sizeof(char));
    char * token = strtok(copy, " ");
    strcpy(temp[0], token); 
    int i = 1;
    while (token && (token = strtok(NULL, " "))) {
        strcpy(temp[i], token);
        i++;
    }
    char * new_null;
    //new_null = NULL; 
    //strcpy(temp[i], new_null);
    if(!temp[i]) printf("yup\n");
    int c = 0;
    for ( ; c <= i; c++) {
        printf("%s ", temp[c]);
    }
    return temp;
}

I am trying to read in a string, space separated, similar to find ./ -name *.h. I am trying to input them into execv.

char (* arguments)[16] = (char **) malloc(256 * sizeof(char));

//...numerous lines of unrelated code

pid = fork();
if (pid == 0) {
    arguments = get_arguments(input_string);
    char * para[] = {"find", "./","-name", "*.h", NULL};
    execv("/usr/bin/find", (char * const *) arguments);
    //printf("%s\n", arguments[0]);
    printf("\nexec failed: %s\n", strerror(errno)); //ls -l -R
    exit(-1);
}

When I swap arguments in the execv call for para it works as intended, but trying to call with arguments returns exec failed: Bad address. If I remove the NULL from para I get the same issue. I've tried strcpy(temp, (char *) NULL), the version you see commented out in get_arguments, and a number of other things that I can't recall in their entirety, and my program ranges from Segmentation fault to failure to compile from attempting to strcpy NULL.

Changing the declarations of arguments and temp to char ** arguments = (char *) malloc(256 * sizeof(char)); ``char ** temp = (char *) malloc(256 * sizeof(char));clears upwarning: initialization from incompatible pointer typebut causes segfault on all calls toget_arguments`.

I am most confused because most places I read say that unused space in a declared array space is defaulted to null, but it fails the if(!temp[i]) printf("yup\n"); test under all conditions I have attempted. — Kebtiz
– Kebtiz, Commented Oct 5, 2016 at 1:33
You can remove the (char *) cast in front of that malloc(), for one thing. — ad absurdum
– ad absurdum, Commented Oct 5, 2016 at 1:41
Thank you, that cleared up warning: initialization from incompatible pointer type. Odd that is a convention I've seen used repeatedly. Are there situations where that explicit type is required? — Kebtiz
– Kebtiz, Commented Oct 5, 2016 at 1:54
Not in C. Some people are very adamant that you should never cast a malloc(), others think that it is OK when it makes your code clearer. You were getting a warning because you were casting the return value of malloc() to char *, but assigning that value to char **. Check out this link: stackoverflow.com/questions/605845/… — ad absurdum
– ad absurdum, Commented Oct 5, 2016 at 2:01

David Schwartz · Accepted Answer · 2016-10-05 01:29:12Z

1

You want this:

char* temp[256]; // an array of 256 char*'s
char * token = strtok(copy, " ");
temp[0] = strdup(token);
int i = 1;
while (token && (token = strtok(NULL, " "))) {
    temp[i] = strdup(token);
    i++;
}
temp[i] = NULL;

answered Oct 5, 2016 at 1:29

David Schwartz

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