How much storage does this Swift struct actually use?

Question

Suppose I have the following struct –

struct MyStruct {
    var value1: UInt16
    var value2: UInt16
}

And I use this struct somewhere in my code like so -

var s = MyStruct(value1: UInt16(0), value2: UInt16(0))

I know that the struct will require 32-bits of storage for the two 16-bit integers –

What I am not certain about is whether swift is allocating two additional 64-bit pointers for each value in addition to one 64-bit pointer for the variable s.

Does this mean total storage requirement for the above code would result in the following?

MyStruct.value1     - 16-bits
MyStruct.value1 ptr - 64-bits
MyStruct.value2     - 16-bits
MyStruct.value2 ptr - 64-bits
s ptr               - 64-bits
–––––––––––––––––––––––––––––
Total               - 224-bits

Can someone please clarify?

Answer 1

MyStruct is 4 bytes because sizeof(UInt16) is 2 bytes. To test this for any given type, use sizeof. sizeof return the memory in bytes.

let size = sizeof(MyStruct) //4

If you want to get the size of a given instance you can use sizeOfValue.

var s = MyStruct(value1: UInt16(0), value2: UInt16(0))
let sSize = sizeofValue(s) //4

I believe the size of the pointer will depend on the architecture/compiler which is 64-bits on most computers and many newer phones but older ones might be 32 bit.

I don't think there is a way to actually get a pointer to MyStruct.value1, correct me if i'm wrong (i'm trying &s.value1.

Pointers

Structs in Swift are created and passed around on the stack, that's why they have value semantics instead of reference semantics.

When a struct is created in a function, it is stored on the stack so it's memory is freed up at the end of the function. It's reference is just an offset from the Stack Pointer or Frame Pointer.