Why Does Incrementing an Integer Pointer in C Increase the Address by 4 Bytes?

Why int Pointer Increment Affects Address by 4 Bytes

In C programming, when the value of an int pointer is incremented by 1, it increases by 4 bytes instead of the expected 1 byte. This is because a pointer variable stores the memory address of a variable, and on most architectures, the integer size is 4 bytes.

Incrementing Pointer by 1 Increments Address by 4

When an int pointer is incremented by 1, it does not move the pointer just by the size of one byte (which is the size of a character). Instead, it moves it by the size of the data type it points to (in this case, an int), which for most architectures is 4 bytes.

Justification for Address Increment by 4

The reason for this increment is to maintain proper alignment when accessing data. Incrementing by 4 bytes ensures that the pointer always points to the start of a memory location that can hold an int value. If the pointer were to increment by only 1 byte, it could potentially point to a memory location that is not aligned correctly, leading to errors or unpredictable behavior.

Visiting the 4 Bytes of an int

If you need to visit the 4 bytes of an int one by one, you can cast the int pointer to a char pointer and then use array indexing. Each byte of the int can be accessed as an element of the char array.

Example:

int a = 1;
int *ptr = &a;

char *cptr = (char *)ptr;
for (int i = 0; i < sizeof(int); i++) {
    printf("%d\n", cptr[i]);
}

This will print the four bytes of a one by one as:

1
0
0
0

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