Why Do Structs with Identical Fields but Different Orders Have Different Sizes in Go?-Golang-php.cn

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Why Do Structs with Identical Fields but Different Orders Have Different Sizes in Go?

Linda Hamilton

Dec 26, 2024 am 03:29 AM

Why Do Structs with Identical Fields but Different Orders Have Different Sizes in Go?

Different Struct Sizes with Varying Field Order

In the provided code, two structs, A and B, with identical fields but different field orders, demonstrate differing sizes. Here's why:

Implicit Padding

Memory alignment requirements for data types dictate that addresses of fields must be multiples of specific values. For int64 fields, this multiple is 8 bytes.

In struct A, the first field is a bool, which takes 1 byte. To align the subsequent int64 field on an 8-byte boundary, 7 bytes of implicit padding are added after a.

In struct B, the first field is b of type int64. Since it's already aligned, only 3 bytes of implicit padding are needed after a to align the following int field, which takes 4 bytes.

Size Implications

Due to this implicit padding, the size of A is 24 bytes (1 byte for a, 8 bytes for b, and 15 bytes of padding). Meanwhile, B is 16 bytes (1 byte for a, 3 bytes of padding, 8 bytes for b, and 4 bytes for c).

Zero-Size Structs

Struct C is declared entirely empty, resulting in a size of 0 bytes. According to the Go language specification, structures with no fields larger than zero have a size of zero.

Addressability and Memory Allocation

For zero-size values, the language allows for the same memory address to be reused for distinct variables. This means that for a := C{}, no memory is actually allocated by the system.

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