freeing an double pointer

Question

int size_of_daten = 5;
char *data[size_of_daten];
char *normal_Pointer = (char*)malloc(sizeof(char) * 100);
int i;

for(i=0; i<size_of_daten; i++) {
    data[i] = (char*)malloc(sizeof(char) * 100);
}

data[0] = "0";
data[1] = "1";
data[2] = "2";
data[3] = "3";
data[4] = "4";

printf("data[2] %s\n",data[2]);

strcpy(normal_Pointer,data[2]);

for(i=0; i<size_of_daten; i++) {
   free(data[i]);
}

free(data);

I just tried this... even I freed the array as I malloced it... also I copied the value of data[2] and didn't point at it... so that shouldn't be the problem...

Answer 1

You're trying to copy the strings "0" , "1" , etc. into your data array: you can't just use = to copy strings, you'll need to use a string library method like strcpy .

Once you assign to your array elements those literal strings, eg: data[0]= "0"; , the array elements no longer point to the memory that you've allocated, they point to memory that doesn't belong to you, and you can't use free . You've lost the references to your memory blocks from malloc , causing a memory leak.

Furthermore, you can't do free(data); because it wasn't allocated using malloc : it's an array allocated on the stack.

Answer 2

This is not how you copy data into an array of char s you've previously allocated:

data[0]= "0";
data[1]= "1";
data[2]= "2";
data[3]= "3";
data[4]= "4";

You're overwriting the pointers with the address of 5 new pointers. At this point you've lost the address of the memory you had allocated, and when you call free you're calling it on the statically allocated string "0" .

You need to use strcpy to copy bytes from one character array into another:

strcpy(dest[0], "0");
strcpy(dest[1], "1");
/* etc */

Answer 3

With data[0]= "0" you are overwriting whatever malloced address returned by malloc . What you should be doing instead is

strcpy(data[0], "0");
strcpy(data[1], "1");
...

Answer 4

First, assignment to the elements of data after allocating leaks the previously allocated memory. When you assign a char *, you assign only the char *: it doesn't copy the string, it overwrites the pointer.

Second, you're freeing data declared on the stack. You define data with:

char *data[size_of_daten];

This describes an on-stack array of character pointers. It will go out of scope when your function returns, and freeing it manually will blow up.