#Pointer problem >~>

34 messages · Page 1 of 1 (latest)

nocturne pagoda
# 
long return_pointer_adress(long* pointer)
{
    long* newPointer = pointer - 2;
    long result = *newPointer;
    return result;
}```
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used pointer arithmetic to get the address smaller by 2 * sizeof(long)

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so im assuming you want to go back by 2 * sizeof(long)

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then we dereference the new pointer to get the value at that address

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nope, it doesnt work like that

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the compiler already knows you are returning by sizeof(long), because it points to long datatype

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to simplify things, the language already knows you are returning by 2

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so no need to add sizeof(long)

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// smaller by 2 * sizeof(long) than this one using pointer-specific operators. this is a trick question, ig

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ye

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as simple as that

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ye

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with n an integer, n * sizeof(long) eq to pointer - n

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nope,

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but whatever the size of the datatype you are using

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the language knows it should go back by n * sizeof that datatype

# 
    long* newPointer = pointer - 2;
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because its already declared as long pointer

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ask

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why would you look for the hard way lol

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well

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sizeof(char) is 1, so it assumes the pointer is pointing to a datatype which size is 1

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therefore you have to write 2 * sizeof(char)

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he will, that code looks like some unsafe code even tho its safe

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the size of char datatype is 1

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but size of long > 1

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prob 8 afaik

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when working with bytes, we use char because it points to 1 size byte

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thats why you see they cast it to (char*), char pointer

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because if you do, you are allowed to use 2 * sizeof(char)

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but stick to the easy way, AI does not know wether the user is a beginner or an operating system developer 😛

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no problem, glad i helped!

hoary tide
#

the compiler adapts the increment value of a pointer according to the size of the type it points to

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

int main(void){
  int* p = malloc(1024);
  printf("%p\n", p);
  printf("%p\n", p+1);
  return 0;
}

on most machines, the addresses will differ by 4