Thread: need help with assigning memory and changing memory addresses

Hey i was watching some stanford lectures on youtube and i like the bit about working with memory. So i figured i would try and do something like passing the address of an array to a function and manually going through the memory and overiding whats in the memory at a certain point to change the array, but im getting confused

Code:

#include <stdio.h>

int changearray(int *arr){
//pass the reference to the first element of array

//*arr should now point to the first elemet of the array?  
//the above doesnt seem to be true but i thought thats what would happen?

//move two elements down the array

int *finaladdress = arr + sizeof(int)*4;

//above im trying to assign finaladdress to point to the 4th element of the passed array
//again this doesnt seem to be working i must be confused. 

//now i want to change the 4th element of the array
finaladdress = 1;

return 0;

}

int main(){

int arr[10];
changearray(&arr);

return 0;


}

i just wrote this quickly and isnt exactly the same code, but it was the same method i was using, i just want to understand how to pass an address then in a seperate function assign a new value to that address.

You don't want sizeof when doing pointer arithmetic. You automatically move down in size-of-thing-being-pointed-at chunks, so if you want to move to the fourth element you would do

Code:

int *finaladdress = arr+3;

EDIT: And &arr is not what you want to pass. You want to pass "arr".

Try this:

Code:

#include <stdio.h>

void changearray(int *arr){
//pass the reference to the first element of array

//*arr should now point to the first elemet of the array?  
//the above doesnt seem to be true but i thought thats what would happen?

//move two elements down the array

int *finaladdress = arr + 4;

//above im trying to assign finaladdress to point to the 4th element of the passed array
//again this doesnt seem to be working i must be confused. 

//now i want to change the 4th element of the array
*finaladdress = 1;
}

int main(){
int i;
int arr[10]= { 0 };
changearray(arr);
printf("\n\n");
for(i = 0; i < 10; i++)
  printf("\nElement %d: %d", i, arr[i]);

return 0;
}

Adjust the +'s as needed.

EDIT: And &arr is not what you want to pass. You want to pass "arr".

I believe for a stack array they are equivalent. The address of arr IS arr (arr = &arr = &arr[0]).

Obviously that's not true for dynamic arrays, since &ptr would be the address of the pointer.

It has the same value, but the wrong type. And since of course people are compiling with warnings turned on high, we might as well get the type right.

ok thanks for your help. Something im confused about.

when i want to set the address of finaladdress i am doing

int *finaladdress = arr + 4;

i assume the address of finaddress is the same as the 4th element of the array so doing this we have changed the address of final address.

then to change the value

*finaladdress = 1 ;

wouldnt this just change the address again?

my vision of how things work are

*finaladdress = //this is basically changing the address finaladdress is stored in
finaladdress = // this stores the value to be associated with that address??

also when changing the address we do arr + 4 do we only do this because the compiler reconises its an int and thus 4 bytes long?

if we were using a void * would we have to use the sizeof(int) to calculate the address?

sorry another question
when i do this

int *finaladdress = arr + 4;
printf("The address of finaladdress: %p", &finaladdress);

it does not equal the same address as any elements of the array.

ok, so pointers all have a unique memory address and their content is just the address of what its pointing too?

if thats the case why do we use pointers? isnt that the way a normal variable works? it has its own address and it contains a value?

sorry for all the questions, im just confused about this.

Yes, a pointer contains an address, (what it's pointing to).

Pointers have a greater utility than any other variable. Computers rely on memory addresses, so pointers are a natural thing to have and use.