Memory alignment

Learn memory alignment and struct packing.

Memory alignment

Memory alignment ensures a piece of data sits at an address that is a multiple of a specific value. The CPU reads aligned data in a single instruction; unaligned data needs multiple accesses and is slower.

_Alignof(int) gives the required alignment of a type; on Linux x86-64 it is 4 for int. In a struct, field order affects the total size: in struct { char tag; int value; } 3 bytes of padding are added after tag; offsetof(struct Item, value) returns 4. Ordering fields from largest to smallest reduces padding.

Item struct

You declare struct Item { char tag; int value; }. tag is 1 byte, value is 4-byte aligned. 3 bytes of padding go in between; offsetof(struct Item, value) returns 4.

In this lesson we print the int alignment (4) with _Alignof(int) and the offset of value (4) with offsetof(struct Item, value). Both are 4. The <stddef.h> header provides the offsetof macro. This knowledge is the foundation for understanding struct size and memory layout.

Exercise Alignment and offset

Declare struct Item { char tag; int value; };. In main print printf("hizalama: %zu\n", (size_t)_Alignof(int)); and printf("ofset: %zu\n", offsetof(struct Item, value));. Add the <stddef.h> header.

Show example solution

#include <stdio.h>
#include <stddef.h>

struct Item {
    char tag;
    int value;
};

int main(void) {
    printf("hizalama: %zu\n", (size_t)_Alignof(int));
    printf("ofset: %zu\n", offsetof(struct Item, value));
    return 0;
}

Test yourself

Match alignment concepts

_Alignof

Type alignment
offsetof

Field offset
padding

Padding byte
struct

Field order
What is _Alignof(int) on Linux x86-64?
- 4
- 1
- 2
- 8
Show answer

4 — int is 4-byte aligned; _Alignof returns 4.

_Alignof call

printf("hizalama: %zu
", (size_t)_Alignof(____));

Show answer

printf("hizalama: %zu
", (size_t)_Alignof(int));

Next lesson: Summary - Performance

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