macOS understanding double pointers in C

[NRose8989]

I having a lot of trouble trying to figure out this problem. Take a look at this function.

void function3(int **twod)
{
	printf("twod is: %p\n", twod); // this prints the location of twod
	printf("*twod + 1 is: %p\n", *twod + 1); 	
}

this function will be passed a 2D array of ints but I really can't figure out what is happening because the first line prints out 0xbffffa54 but the next line will print out 0x6, which is the value at twod[0][2].

Here is the array as it is declared in main:

int twod[4][3] = {{2,4,6}, {8,10,12}, {14, 16, 18}, {20, 22, 24}};

Can someone please explain to me what is going to with *twod + 1? because naturally I would think that it first dereferenced twod then adds 1 to where it points to..... but now I'm really confused. thanks

 

[lee1210]

Unless you did a little doing to twod in main, things aren't going to go well passing it to function3. Here's the code I wrote to do something similar to this two different ways:

#include <stdio.h>

void function3(int **);
void function4(int twod[][3]);

int main(int argc, char *argv[]) {
  int twod[4][3] = {{2,4,6}, {8,10,12}, {14, 16, 18}, {20, 22, 24}};
  int *twodPass[4];
  twodPass[0]=twod[0];
  twodPass[1]=twod[1];
  twodPass[2]=twod[2];
  twodPass[3]=twod[3];
  function3(twodPass);
  function4(twod);
  return 0;
}

void function3(int **twod) {
        printf("twod is: %p\n", twod); // this prints the location of twod
        printf("*twod + 1 is: %p\n", *twod + 1); 
}

void function4(int twod[][3]) {
        printf("twod is: %p\n", twod); // this prints the location of twod
        printf("*twod + 1 is: %p\n", *twod + 1); 
}

Output:
twod is: 0xbffffc80
*twod + 1 is: 0xbffffc94
twod is: 0xbffffc90
*twod + 1 is: 0xbffffc94

An int ** is different from an int[x][y]. One is a pointer to a pointer, or perhaps a pointer to an array of pointers. The other is x*y*sizeof(int) bytes that you can access conveniently with two indexes. In the example i set up an int *[] to allow for function3 to operate properly. You didn't post all of your code, but i think it would be fair to guess that there was a warning. EDIT: I'm a bit tired, and should be sleeping. Gruffalo's explanation of why you got 0x6 is correct. You can test by modifying the first value of twod.

-Lee

 

[Gruffalo]

I having a lot of trouble trying to figure out this problem. Take a look at this function.

void function3(int **twod)
{
	printf("twod is: %p\n", twod); // this prints the location of twod
	printf("*twod + 1 is: %p\n", *twod + 1); 	
}

this function will be passed a 2D array of ints but I really can't figure out what is happening because the first line prints out 0xbffffa54 but the next line will print out 0x6, which is the value at twod[0][2].

Here is the array as it is declared in main:

int twod[4][3] = {{2,4,6}, {8,10,12}, {14, 16, 18}, {20, 22, 24}};

Can someone please explain to me what is going to with *twod + 1? because naturally I would think that it first dereferenced twod then adds 1 to where it points to..... but now I'm really confused. thanks

It is probably easier to first point out that the declaration of the 2D array is just a convenience. In memory, you get from exactly the same as you would get if you declared:

int oned[12] = {2,4,6,8,10.....};

So, your function3 is expecting an array of pointers to integers. When you dereference that you are saying 'what is the value' of the first pointer in the array, and the first 'pointer', has the value or address 0x2. Now if you increment an integer pointer by 1, you are adding 4 (4 bytes in an integer) which gives you the answer 0x6.

 

[NRose8989]

It is probably easier to first point out that the declaration of the 2D array is just a convenience. In memory, you get from exactly the same as you would get if you declared:

int oned[12] = {2,4,6,8,10.....};

So, your function3 is expecting an array of pointers to integers. When you dereference that you are saying 'what is the value' of the first pointer in the array, and the first 'pointer', has the value or address 0x2. Now if you increment an integer pointer by 1, you are adding 4 (4 bytes in an integer) which gives you the answer 0x6.

OK now that makes sense to me. thanks.

 

[rrpalma]

to the OP, please realize that assuming p is a pointer to an int,
*p + 1
adds 1 to the value pointed by p
whereas *(p+1)
dereferences the integer right next to the one pointed by p (which is probably what you wanted)

Perhaps the following quick C++ code might help a little (or further confuse you):

#include <iostream>
using namespace std;

void f1(int (*p)[3]);
void f2(int *q);

int main() {

	int myArray[4][3] = { { 2, 4, 6 }, { 8, 10, 12 }, { 14, 16, 18 }, { 20,
			22, 24 } };
	cout << "Program starting" << endl;
	f1(myArray);
	f2(myArray[0]);
	return 0;
}

void f1(int (*p)[3]) {
	cout << "pointer p is " << p << endl;
	cout << "item pointed to by p is " << **p << endl;
	cout << "item pointed to by p + 2 is " << **(p+2) << endl;
	return;
}

void f2(int *q) {
	cout << "pointer q is " << q << endl;
	cout << "item pointed to by q is " << *q << endl;
	cout << "item pointed to by q + 2 is " << *(q+2) << endl;
	return;
}

the above code produces the following output on Eclipse Ganymede, Linux 64 bit, with CDT:

Program starting
pointer p is 0x7fff19d23820
item pointed to by p is 2
item pointed to by p + 2 is 14
pointer q is 0x7fff19d23820
item pointed to by q is 2
item pointed to by q + 2 is 6