Why Does Incrementing an Integer Pointer Add 4 Bytes Instead of 1?

Incrementing an Int Pointer: Why the 4-Byte Increment?

When working with pointers, it may be intuitive to expect that incrementing an int pointer would advance its value by 1 byte, as occurs with arrays. However, in practice, incrementing an int pointer actually increases its value by 4 bytes.

The Reason for 4-Byte Increment

The reason for this behavior lies in the size of an integer data type. An int typically occupies 4 bytes of memory. If an int pointer were to increment by only 1 byte, it would result in pointing to a partial integer, which would be nonsensical.

Understanding the Pointer Increment

To illustrate, consider the following memory representation:

[...| 0 1 2 3 | 0 1 2 3 | ...]
[...| int     | int     | ...]

Here, each int occupies 4 bytes. When an int pointer increments by 1, it logically makes sense for it to move to the next 4-byte section, as shown below:

             [↓      ]
[...| 0 1 2 3 | 0 1 2 3 | ...]
[...| int     | int     | ...]

This ensures that the pointer continues to point to a valid integer.

Accessing Individual Bytes

If the need arises to access individual bytes of an integer, you can utilize a char pointer. Since char always has a size of 1 byte, you can use a char* pointer to increment by one byte at a time and access the corresponding bytes of an integer.

Example:

int i = 0;
int* p = &i;

char* c = (char*)p;
char x = c[1]; // access the second byte of the integer

Additional Note:

It's important to remember that incrementing a void* pointer is not allowed, as void is an incomplete type and has no defined size.

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