Thread: using free()

You have to do the reverse of the allocation. Loop through freeing each element and then free the array of pointers.

Don't mean to be rude here, but why are you even doing it this way? Why not use calloc() to allocate a multi-dimensional array?

But speaking of your problem:

I am pretty sure that what is happening is that your overwrite your first pointer (cptr) with another pointer cptr[0], who both point to the same place, and then you are trying to free the same address twice. Hope this make sense...

but why are you even doing it this way? Why not use calloc() to allocate a multi-dimensional array?

malloc and calloc do the same exact thing:

calloc( sizeof( somevar ), number_of_elements );

Is the same as:

malloc( sizeof( somevar ) * number_of_elements );

The end result is the same thing. You gain nothing by calling 'calloc' over 'malloc', other than typing ',' instead of '*', and 'c' instead of 'm'.

In this case, 'calloc' will not help you. If you want randomly sized strings, you still have to allocate them seperately:

myStrings = calloc( sizeof( char* ), number_of_pointers );
myStrings = malloc( sizeof( char* ) * number_of_pointers );

They're the same thing.

Quzah.

I beg to differ...

malloc() only takes the length of the memory to allocate:

malloc(sizeof(int));

when calloc() takes the number of elements to allocate and the length of individual element:

calloc( 10/*elements*/, sizeof(int));

I beg to differ...

malloc() only takes the length of the memory to allocate:

Like I said, malloc and calloc work the same. The end result is exactly the same:

x = calloc( num, size( x ) );
x = malloc( num * size(x) );

for( y=0; y<num; y++ ) x[y] ...

In either case, the end result is EXACTLY the same.

Quzah.

And both are true...

void* calloc (size_t sizeObjCnt, size_t sizeObject);
void* malloc (size_t sizeObject);









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If I might humbly point out, space allocated by calloc will be initialized to 0. Space allocated by malloc is uninitialized.

What you've posted seems ok (cptr[0] doesn't point to the same location as cptr). Make sure that the number of allocations match. You've allocated five pointers in your first post, does MAX_ENTRY equal five? The complete alloc/dealloc should look something like this -

Code:

#include <stdlib.h>

#define MAX_ENTRY 5

int main() {

	int ctr;
	char **cptr; 
	cptr = malloc(sizeof(char *) * MAX_ENTRY); 
	for (ctr = 0; ctr < MAX_ENTRY; ctr++) 
	{ 
		cptr[ctr] = malloc(20); 
 
	} 

	//Do stuff

	for (ctr = 0; ctr < MAX_ENTRY; ctr++) 
	{ 
		free(cptr[ctr]); 
	} 
	
	free(cptr);
    return 0;
}

7.14.3.1 The calloc function

Synopsis

[#1]

#include <stdlib.h>
void *calloc(size_t nmemb, size_t size);

Description

[#2] The calloc function allocates space for an array of
nmemb objects, each of whose size is size. The space is
initialized to all bits zero.225

Returns

[#3] The calloc function returns either a null pointer or a
pointer to the allocated space.

7.14.3.3 The malloc function

Synopsis

[#1]

#include <stdlib.h>
void *malloc(size_t size);

Description

[#2] The malloc function allocates space for an object whose
size is specified by size and whose value is indeterminate.

Returns

[#3] The malloc function returns either a null pointer or a
pointer to the allocated space.


That should clear up and arguments over malloc() vs. calloc().... also all bits 0 isn't guaranteed to mean it holds the actual value of 0.

You're missing the entire point. The point was:

Don't mean to be rude here, but why are you even doing it this way? Why not use calloc() to allocate a multi-dimensional array?

When you come in and say "Why are you even doin it this way?" and "Why not use calloc()..." it implies that 'calloc' is the better choice. In reality, using malloc and using calloc have the same thing. Furthermore, you could even use 'realloc' and end up with the same end result. That was my point. They all produce the same endresult: an allocated block of memory.

In other words: They _ALL_ will produce "an array" of allocated data if you want them to. Furthermore, calloc an do exactly what malloc does if you tell it to only allocate 1 member. That is my point. They all function the same, there is no "wrong" way here.

Quzah.

Um... I don't think I missed any point. I wasn't trying to prove or disprove anyone, I was just posting what the standard said about the functions to clear up and misconceptions people had about them. Quzah you seem (to me, though I am probably very wrong) to think that I am trying to target you in most posts when actually I agree with most things you say... maybe you are reading too deep into things?

Sorry. I jump the gun some times. I was just saying that I wasn't confused on how they work, and that was my whole point.

Quzah.