Thread: using malloc for an array of character pointers

I am able to work with n instances of a structure in one mallocated area, but when I try to do the same thing with just character pointers, I get compiler errors about making integer from pointer without a cast. If I create a structure with just a character pointer in it, it works just fine... I am just not seeing something here!!!

This works:

Code:

#include <stdio.h>
#include <stdlib.h>


int main (void)
{
    struct items
    {
        unsigned int item_1;
        unsigned int item_2;
    };
    struct items *items_ptr;
unsignedint i = 0;
unsignedint num = 0;
    items_ptr = (struct items *) malloc (3 * sizeof(struct items));
    if (items_ptr == NULL)
    {
            printf ("\nOut of memory - ending program.");
return1;
    }
    for (i = 0; i < 3; i++)
    {
        num = num + 1;
        items_ptr[i].item_1 = num;
        items_ptr[i].item_2 = num + 10;
    }
    for (i = 0; i < 3; i++)
    {
        printf ("%u %u \n", items_ptr[i].item_1, items_ptr[i].item_2);
    }
    free (items_ptr);
    items_ptr = NULL;
    return 0;
}

This DOES NOT work!

Code:

#include <stdio.h>
#include <stdlib.h>


int main (void)
{
    char * items_ptr = NULL;
unsignedint i = 0;
    char * one = "one";
    char * two = "two";
    char * three = "three";
    items_ptr = (char *) malloc (3 * sizeof(char *));
    if (items_ptr == NULL)
    {
            printf ("\nOut of memory - ending program.");
return1;
    }
    items_ptr[0] = (char *) one;
    items_ptr[1] = two;
    items_ptr[2] = three;
    for (i = 0; i < 3; i++)
    {
        printf ("%s", items_ptr[i]);
    }
    free (items_ptr);
    items_ptr = NULL;
    return 0;
}

you want a pointer to a char *

Code:

char ** items_ptr = NULL;

Kurt

That gives me a warning about assignment from incompatible pointer type.

Originally Posted by dtkokovoko

That gives me a warning about assignment from incompatible pointer type.

it is no good idea to cast the return of malloc.
Kurt

Sorry! That was an experimentation relic. Once removed, your suggestion worked perfectly!!! I will have to dig deeper to understand what and why...

Many thanks!!!

Originally Posted by dtkokovoko

That gives me a warning about assignment from incompatible pointer type.

The code is now as shown below. Interestingly, any time I edit it, the first time I do a build, I get the warning above. The second time I do the build it completes without the warning.

Is this normal?

Code:

#include <stdio.h>
#include <stdlib.h>


int main (void)
{
	char ** items_ptr = NULL;
unsignedint i = 0;
	char * one = "one";
	char * two = "two";
	char * three = "three";
	items_ptr = (char *) malloc (3 * sizeof(char *));
	if (items_ptr == NULL)
	{
		printf ("\nOut of memory - ending program.");
return1;
	}
	items_ptr[0] = one;
	items_ptr[1] = two;
	items_ptr[2] = three;
	for (i = 0; i < 3; i++)
	{
		printf ("%s\n", items_ptr[i]);
	}
	free (items_ptr);
	items_ptr = NULL;
    return 0;
}

Originally Posted by ledow

Did you not read the reply?

Code:

char ** items_ptr

Code:

items_ptr = (char *) malloc

Do not cast malloc or, if you do, make sure it's cast to the same type as the variable you're assigning to.

Yes, I read the reply.

I thought he was referring to the "experimentation relic", the (char *) cast attempt, on the first version.

Code:

    items_ptr[0] = (char *) one;

I thought he meant I just needed to change the declaration to char **.

I am still working on figuring out what char ** means, but, so far, I have not found it in my reference books, and my online search capability goes stupid when I include ** in a search. So far my understanding is that, as he said, what I need is a pointer to a char * (actually to an array of char), (is that what a char ** is? - I have no idea!) but I also thought that the cast before malloc related to how the allocated memory was parsed.

Thank you for the next hint: both sides of the assignment need to be the same type.

Thank you for the explanation, and thank all of you for your guidance!!!