Dynamically grown array of strings

Question

I'm trying to build a dynamically grown array of strings. Neither the number of strings nor the length of each string is known at compile time. Here's the code I came up with so far (this is just me playing with the syntax):

char **result = NULL;
char *temp = NULL;
result = (char **)realloc (result,sizeof(char *) * 1);
temp= (char *)realloc(temp,5 * sizeof(char));

strcat(temp,"hello");


temp= (char *)realloc(temp,10 * sizeof(char));

strcat(temp," world");

printf ("%s \n", temp);
result[0]=temp;
free(temp);
printf ("%s \n", result[0]);

result = (char **)realloc (result, sizeof(char *) * 2);
temp= (char *)realloc(temp,10 * sizeof(char));
strcat(temp,"0123456789");

temp= (char *)realloc(temp,15 * sizeof(char));
strcat(temp,"asdfg");

printf ("%s \n", temp);
result[1]=temp;
free(temp);
printf ("%s \n", result[0]);
printf ("%s \n", result[1]);)

Now, when I print result[0] or result[1], its just an empty string, why doesn't result[1]=temp; work?

Here's what I tried earlier, but it didn't work, I kept getting "invalid size" errors when using realloc() on that last line:

char **result = NULL;
result = (char **)realloc (result,sizeof(char *) * 1);
result[0]= (char *)realloc(result[0],5 * sizeof(char));

strcat(result[0],"hello");
printf ("%s \n", result[0]);

result[0]= (char *)realloc(result[0],10 * sizeof(char));

strcat(result[0]," world");
printf ("%s \n", result[0]);

result = (char **)realloc (result, sizeof(char *) * 2);
result[1]= (char *)realloc(result[1],10 * sizeof(char));
strcat(result[0],"0123456789");

result[0]= (char *)realloc(result[1],15 * sizeof(char));
strcat(result[0],"asdfg");

If anybody could help me get either version working, I would be very grateful.

UPDATE: Ok, I got both versions of the code working. Now when I try to use this same format in my actual program, I get errors such as

*** glibc detected *** ./uvicfmt3: realloc(): invalid next size: 0x08ed3170 ***

Now in my program, the "result" is declared as a global variable (using the second version of my code), and the realloc functions are called in different subroutines. Is that what's causing the problem? How could I get around this?

The only function for the casts is code obfuscation. Make sure you have #include <stdlib.h> and get rid of the casts. If you're compiling with a C++ compiler, try to write C++ in the 1st place ;) — pmg
– pmg, Commented Nov 30, 2011 at 15:01
that error realloc(): invalid next size - really, any error from *alloc besides "out of memory" - is a sign. It means you've probably corrupted the heap, overwriting the data structures that *alloc uses to keep track of the location and size of free blocks. When you see that error, start looking for invalid array writes, double frees, and stuff like that. — japreiss
– japreiss, Commented Nov 30, 2011 at 16:30
"hello" takes 6 bytes (5 for hello + 1 for terminating ASCII NUL), "hello world" takes 12 bytes. — ninjalj
– ninjalj, Commented Nov 30, 2011 at 19:00

NPE · Accepted Answer · 2011-11-30 14:21:17Z

2

The following statement makes both result[0] and temp point to the same memory address:

result[0]=temp;

Having performed the above assignment, you then free(temp) and try to access result[0]:

free(temp);
printf ("%s \n", result[0]);

This is undefined behaviour, since you're accessing the memory that's just been deallocated.

The same goes for the identical code you have for result[1].

edited Nov 30, 2011 at 14:21

answered Nov 30, 2011 at 14:15

NPE

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cnicutar · Accepted Answer · 2011-11-30 18:18:13Z

2

In the first example you are using a string after freeing it
In the second example you are strcating to unallocated memory (plus the uneducated realloc(result[0] makes no sense)

You could try this:

char **result = NULL;
result = realloc(result, sizeof(char *) * 1);

result[0] = strdup("hello");

/* ... */

result = realloc(result, sizeof(char *) * 2);
result[1] = strdup(" world");

Now strdup isn't standard but it's not hard to steal it / fake it. Here's one attempt:

char *strdup2(const char *str)
{
    size_t len;
    char *rval = NULL;

    len = strlen(str);        /* We should probably check this. */
    rval = malloc(len + 1);   /* And this. */

    memcpy(rval, str, len);
    rval[len] = 0;

    return rval;
}

EDIT

I was under the (possibly wrong) impression you want to later modify the strings. If that is not the case, simply storing them (without strdup) is adequate:

result[0] = "hello";

edited Nov 30, 2011 at 18:18

answered Nov 30, 2011 at 14:17

cnicutar

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2 Comments

user1073407 Over a year ago

Could you please explain why strdup is necessary in this case?

cnicutar Over a year ago

@user1073407 It's only necessary if you plan to modify the strings. Otherwise it's perfectly legitimate to simply store the literals.

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