Malloc and pointers

I got the following code:

Code:

---

#define COLS 5

typedef int RowArray[COLS];
RowArray *rptr;

int main(void)
{
    int nrows = 10;
    int row, col;
    rptr = malloc(nrows * COLS * sizeof(int));
    for (row = 0; row < nrows; row++)
    {
        for (col = 0; col < COLS; col++)
        {
            rptr[row][col] = 17;
        }
    }

    return 0;
}

---

I am confused on this line: rptr = malloc(nrows * COLS * sizeof(int));

Wouldnt that just create a like a long memory storage space? How does the compiler know that it has rows and cols??

thanks

"Wouldnt that just create a like a long memory storage space?"

It would create a space 200 bytes long, which is needed to store the eventual contents of *rptr.

"How does the compiler know that it has rows and cols??"

I don't think it matters whether the compiler can "know this" as long as it can multiply 10 x 5 x 4.

This presumes sizeof(int) is 4, which it is on most systems.

so the compiler would know that the first 10 spaces is one row etc?

A "RowArray" here is an array of five ints. By mallocating enough space to *ptr, you can implicitly make it an array of whatever it points to -- eg, with a normal int:

Code:

---

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

int main() {
        int x, *ptr=&x, i;
        ptr=malloc(12);
        ptr[1]=7;
        for (i=0;i<3;i++) printf("%d\n", ptr[i]);
}

---

So in your example, *rptr ends up as a 2 dimensional int array -- the first number is the element of the pointer array, the second the element of the array pointed to by the element of the pointer array.

what does this mean the void:
compare((void *)&p[j-1]

Im confused on the void *

Quote:

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Originally Posted by taurus

Why is this case dont i use ** for the pointer?

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Sorry, I should have written "*rptr effectively ends up as a 2 dimensional int array". *rptr literally is still just a pointer, but it points to a space in memory containing an array of RowArrays. A RowArray is an array of five ints.

** indicates a true multi-dimensional array, because it's a pointer to a pointer.

Quote:

---

Originally Posted by taurus

what does this mean the void:
compare((void *)&p[j-1]

Im confused on the void *

---

void* casts &p[j-1] as a void pointer.

Why do you do that? is it so that you can then cast back that data to whatever typ you need? something like that

Quote:

---

Originally Posted by taurus

Why do you do that? is it so that you can then cast back that data to whatever typ you need? something like that

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The casting doesn't change &p[j-1], but it does cause "compare" to consider it as a void pointer.