Questions about pointers and malloc

If I have

Code:

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char **array = malloc (sizeof *array);

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What have I created? What's array pointing to?


If I write

Code:

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size_t row; /* defined before calling malloc */
size_t size; /* defined later before calling malloc */
.
.
.
array[row] = malloc (size);
.
.
.
++row;

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How many array[row] = malloc (size); can I write, given that row is increased each time? Is it correct to do that?

Quote:

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Originally Posted by H`eya

What have I created? What's array pointing to?

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A pointer to a pointer. Of whatever type "array" belongs.

Quote:

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How many array[row] = malloc (size); can I write, given that row is increased each time? Is it correct to do that?

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This looks like something you might use this way:

Code:

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int *array[10];
for (row=0;row<10;row++)
  array[row] = malloc(SIZE*sizeof(int))

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in which case you would have an array of ten pointers to arrays of SIZE ints.

Appreciate your prompt response.
Allow me to explain further then, since I think I did not give you enough to clarify my concerns.

Code:

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

int main (void )
{
    size_t i;
    size_t row = 5, size = 32;
    char **array = malloc(sizeof *array);
    if (!array) {
        /* handle the error */
    }

    /*
    * This is the part I am interested in
    * knowing if it would be correct
    */
    for (i = 0, i < row, i++) {
        *array[i] = malloc(size); /* does **array have memory for *array[i]? */
        if (*array == NULL) {
          /* handle the error */
        }
    }

    return 0;
}

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I apologize that I did not explain well in my previous post.

malloc gives you the amount of memory asked for. You ask for enough memory to hold one of "* array" things. So you have enough room for one, not 5.

Many thanks, to both of you.
I understand.