Function std::mem::size_of

pub const fn size_of<T>() -> usize

Returns the size of a type in bytes.

More specifically, this is the offset in bytes between successive elements in an array with that item type including alignment padding. Thus, for any type T and length n, [T; n] has a size of n * size_of::<T>().

In general, the size of a type is not stable across compilations, but specific types such as primitives are.

The following table gives the size for primitives.

Type	size_of::<Type>()
()	0
u8	1
u16	2
u32	4
u64	8
i8	1
i16	2
i32	4
i64	8
f32	4
f64	8
char	4

Furthermore, usize and isize have the same size.

The types *const T, &T, Box<T>, Option<&T>, and Option<Box<T>> all have the same size. If T is Sized, all of those types have the same size as usize.

The mutability of a pointer does not change its size. As such, &T and &mut T have the same size. Likewise for *const T and *mut T.

Examples

use std::mem;

// Some primitives
assert_eq!(4, mem::size_of::<i32>());
assert_eq!(8, mem::size_of::<f64>());
assert_eq!(0, mem::size_of::<()>());

// Some arrays
assert_eq!(8, mem::size_of::<[i32; 2]>());
assert_eq!(12, mem::size_of::<[i32; 3]>());
assert_eq!(0, mem::size_of::<[i32; 0]>());


// Pointer size equality
assert_eq!(mem::size_of::<&i32>(), mem::size_of::<*const i32>());
assert_eq!(mem::size_of::<&i32>(), mem::size_of::<Box<i32>>());
assert_eq!(mem::size_of::<&i32>(), mem::size_of::<Option<&i32>>());
assert_eq!(mem::size_of::<Box<i32>>(), mem::size_of::<Option<Box<i32>>>());