Arrays of Strings. With functions/memcpy?

**Kleid-0** · 01-10-2005

I am super lost. You see when I'm at school I like to write functions about what I want to do when I get home. Well I was creating a sorting function that sorts an array of strings, it doesn't work just yet. Well, there's no C compiler here, and I was wondering if someone could help me out if I have concept problem with what's going on with this code:

Code:

#include <stdio.h>

char **sortAll(const char *sortThis[255], const int nTotal) {
	char *szSorted[255] = malloc(nTotal * sizeof(*szSorted));
	strcpy(szSorted[0], sortThis[0]);
	return szSorted;
}

int main(void) {
	const char *szNotSorted[] = {
		"abc",	"def"
	}
	char **szSorted;
	memcopy(szSorted, sortAll(szNotSorted, 2));
	puts(szSorted[0]); /* Hoping 'abc' :) */
	return 0;
}

I've never used memcopy before lol, and it just looked right at the time. I highlighted the parts where I have no idea in red. Could you tell me what I'm doing wrong? Why it wouldn't compile correctly?

**quzah** · 01-10-2005

Your memory allocation is wrong. You can't simply malloc up a huge block and assign it to every member of an array in one shot. You have to loop through each element and assign them space individually. Furthermore, you shouldn't even bother with an array there. Just use a pointer to a pointer:

Code:

char **foo( const char *data[255], int nblocks )
{
    char **bar;
    int x;

    /* error check nblocks here... */

    /* allocate nblocks worth of pointers */
    bar = malloc( sizeof( char * ) * nblocks );
    if( bar == NULL )
        return NULL; /* handle your error however, I'm doing this... */

    /* allocate space for each string... */
    for( x = 0; x < nblocks; x++ )
        bar[x] = malloc( sizeofyourstringspace );

    /* do stuff, copying or sorting your data */


    return bar;
}

You can actually "fake" sorting this. Instead of allocating space for each string and copying the data into it, simply use the pointers you've allocated and shuffle around what it points at, making them point to the first array's data in a sorted manner.

Also, when you're done here, if in fact you are mallocing for each string, you need to free in the reverse manner in which you allocate:
1) loop through and free each allocated string, one at a time.
2) free the origional pointer chunk allocation

Code:

for( x = 0; x < nblock; x++ )
    free( bar[x] );
free( bar );

Quzah.

**Zach L.** · 01-10-2005

Your spelling is also wrong. There is no 'o' in memcpy.

Very brief reference: http://cppreference.com/stdstring_details.html#memcpy

**Kleid-0** · 01-10-2005

Originally Posted by quzah

Also, when you're done here, if in fact you are mallocing for each string, you need to free in the reverse manner in which you allocate:

Code:

for( x = 0; x < nblock; x++ )
    free( bar[x] );
free( bar );

But wouldn't that need to be:

Code:

for( x = nblock; x >= 0; --x )
    free( bar[x] );
free( bar );

?

**quzah** · 01-10-2005

No. The order you free each element in your block doesn't matter. It's just like an array of pointers which have data allocated for them. It doesn't matter how you move through the array freeing what's there, just so long as you cover every cell. You could skip around randomly if you felt like it.

Quzah.

**Kleid-0** · 01-11-2005

Code:

bar = malloc( sizeof( char * ) * nblocks );
if( bar == NULL )
     return NULL; /* Handle malloc() errors */

How could bar == NULL if it's allocated memory? Could malloc mess up on mistake? I thought malloc was perfect :(

**Stack Overflow** · 01-11-2005

Hello,

According to this reference: If the system could not allocate the requested block of memory or if the size requested was 0 a NULL pointer is returned.

- Stack Overflow

**Kleid-0** · 01-11-2005

I just checked my C reference, you're right. I love C, the more problems I have, the better this language gets! I am truley addicted.

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