How do you Assign an address to an element of pointer array?

Answer 1

Pointers were always difficult subject to understand. I would suggest to read several tutorials and spend some time playing with them.

Explanation:

/* create array of two pointers to integer */

int *aaa[2];

/* setting first pointer in array to point to 0x1212 location in memory */

aaa[0] = (int *)0x1212;

/* allocate memory for the second pointer and set value */

aaa[1] = malloc(sizeof(int));

*aaa[1] = 333;

/* if you want to can make first item to point to the second,

now arr[0] pointer address changed to the same as aaa[1] pointer */

aaa[0] = aaa[1];

/* print aaa[1] address and later print value */

printf("LOCATION: %X\n", (unsigned int)*(aaa + 1));

printf("VALUE: %d\n", **(aaa + 1));

/* exact the same here, but different way of how I wrote pointers */

printf("LOCATION: %X\n", (unsigned int)aaa[1]);

printf("VALUE: %d\n", *aaa[1]);

If you would do the same with 0 element of array, which points to 0x1212 memory location you would get some kind of number or Bus error if you accessed restricted memory place (for example the beginning of memory).

I hope this I did not make you more confused than you are.

Answer 2

You use the assignment operator.

For example:

int * p[10]; // Declare an array of 10 pointers to int.

int x = 42; // Declare an int named x.

p[0]=&x; // Assign the address of x to the first element of the array.

Answer 3

Use the addressof operator (&) on the elements you want to point at.

The following code demonstrates an array of 10 integers and 10 separate pointers, one pointing at each element in the array. As you will see, the pointers don't provide any information you can't get from the array itself; the data in the first two columns is exactly the same as the data in the last two columns. The middle column simply shows the address of each pointer (pointer's are just variables, after all).

#include <iostream>

int main()

{

int arr[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

int * p0 = &arr[ 0 ]; printf( "&arr[0]=0x%.8x\tarr[0]=%.2d\t&p0=0x%.8x\tp0=0x%.8x\t*p0=%.2d\n", &arr[0], arr[0], &p0, p0, *p0 );

int * p1 = &arr[ 1 ]; printf( "&arr[1]=0x%.8x\tarr[1]=%.2d\t&p1=0x%.8x\tp1=0x%.8x\t*p1=%.2d\n", &arr[1], arr[1], &p1, p1, *p1 );

int * p2 = &arr[ 2 ]; printf( "&arr[2]=0x%.8x\tarr[2]=%.2d\t&p2=0x%.8x\tp2=0x%.8x\t*p2=%.2d\n", &arr[2], arr[2], &p2, p2, *p2 );

int * p3 = &arr[ 3 ]; printf( "&arr[3]=0x%.8x\tarr[3]=%.2d\t&p3=0x%.8x\tp3=0x%.8x\t*p3=%.2d\n", &arr[3], arr[3], &p3, p3, *p3 );

int * p4 = &arr[ 4 ]; printf( "&arr[4]=0x%.8x\tarr[4]=%.2d\t&p4=0x%.8x\tp4=0x%.8x\t*p4=%.2d\n", &arr[4], arr[4], &p4, p4, *p4 );

int * p5 = &arr[ 5 ]; printf( "&arr[5]=0x%.8x\tarr[5]=%.2d\t&p5=0x%.8x\tp5=0x%.8x\t*p5=%.2d\n", &arr[5], arr[5], &p5, p5, *p5 );

int * p6 = &arr[ 6 ]; printf( "&arr[6]=0x%.8x\tarr[6]=%.2d\t&p6=0x%.8x\tp6=0x%.8x\t*p6=%.2d\n", &arr[6], arr[6], &p6, p6, *p6 );

int * p7 = &arr[ 7 ]; printf( "&arr[7]=0x%.8x\tarr[7]=%.2d\t&p7=0x%.8x\tp7=0x%.8x\t*p7=%.2d\n", &arr[7], arr[7], &p7, p7, *p7 );

int * p8 = &arr[ 8 ]; printf( "&arr[8]=0x%.8x\tarr[8]=%.2d\t&p8=0x%.8x\tp8=0x%.8x\t*p8=%.2d\n", &arr[8], arr[8], &p8, p8, *p8 );

int * p9 = &arr[ 9 ]; printf( "&arr[9]=0x%.8x\tarr[9]=%.2d\t&p9=0x%.8x\tp9=0x%.8x\t*p9=%.2d\n", &arr[9], arr[9], &p9, p9, *p9 );

return( 0 );

}