Pointer arithmethic

Effects of using different pointer data types on pointer READ

The data type used when declaring a pointer i.e. (char* PointerVariable) determines how much data will be accessed or fetched when dereferencing that pointer, for example, the former pointer will only fetch char --> 1 byte of data, a pointer declared with short will fetch 2 bytes of data.

To illustrate this, execute the following code:

#include<stdio.h>

long long int g_data = 0x8877665544332211;

int main(void){
    char* pAddress1;
    pAddress1 = (char*)&g_data;
    printf("Value at address %p is: %x\n", pAddress1, *pAddress1);

    short *pAddress2;
    pAddress2 = (short*)&g_data;
    printf("Value at address %p is: %x\n", pAddress2, *pAddress2);

    int *pAddress3;
    pAddress3 = (int*)&g_data;
    printf("Value at address %p is: %x\n", pAddress3, *pAddress3);

    long long int *pAddress4;
    pAddress4 = (long long int*)&g_data;
    printf("Value at address %p is: %I64x\n", pAddress4, *pAddress4);
    
    return 0;
    
}

Output

Value at address 00007FF71CCE4010 is: 11
Value at address 00007FF71CCE4010 is: 2211
Value at address 00007FF71CCE4010 is: 44332211
Value at address 00007FF71CCE4010 is: 8877665544332211

where each pointer type fetches only a part of the data stored in the variable g_data and the address remains the same regardless of which data type is chosen for each pointer.

Pointer Arithmetic (Addition)

What happens when you add to a pointer?

The answer lies in how that specific pointer was declared. Like we saw in the previous example, each data type determines how much data will be read/fetched and it will also determine how many bytes will the address jump when adding a integer to a pointer.

To illustrate this, let's add one to a char pointer.

#include<stdio.h>

long long int g_data = 0x8877665544332211;

int main(void){
    char *pAddress = (char*)&g_data;
    printf("value of pAddress %p is %x\n", pAddress, *pAddress);

    pAddress = pAddress + 1;    // Traverse to the next byte

    printf("value of pAddress %p is %x\n", pAddress, *pAddress);
}

Output

value of pAddress 00007FF6B5E64010 is 11
value of pAddress 00007FF6B5E64011 is 22

Where, because the pointer was declared as a char (1 byte), when an addition of one is made, the address changes only one byte.

If we instead we add 5 bytes, the result is the following:

#include<stdio.h>

long long int g_data = 0x8877665544332211;

int main(void){
    char *pAddress = (char*)&g_data;
    printf("value of pAddress %p is %x\n", pAddress, *pAddress);

    pAddress = pAddress + 5;    // Traverse to the 6th byte

    printf("value of pAddress %p is %x\n", pAddress, *pAddress);
}

Output

value of pAddress 00007FF6CC5A4010 is 11
value of pAddress 00007FF6CC5A4015 is 66

Where we traversed 5 bytes.

Similarly, if we want to traverse the data 4 bytes at at time, we can use an int with the GCC compiler:

#include<stdio.h>

long long int g_data = 0x8877665544332211;

int main(void){
    int *pAddress = (int*)&g_data;

    printf("value of pAddress %p is %x\n", pAddress, *pAddress);

    pAddress = pAddress + 1;

    printf("value of pAddress %p is %x\n", pAddress, *pAddress);
}

Output

value of pAddress 00007FF6171D4010 is 44332211
value of pAddress 00007FF6171D4014 is 88776655

Where we see that we need to sum only one to the pointer to traverse all of the data.