Function std::mem::transmute_copy

pub const unsafe fn transmute_copy<Src, Dst>(src: &Src) -> Dst

Interprets src as having type &Dst, and then reads src without moving the contained value.

This function will unsafely assume the pointer src is valid for size_of::<Dst> bytes by transmuting &Src to &Dst and then reading the &Dst (except that this is done in a way that is correct even when &Dst has stricter alignment requirements than &Src). It will also unsafely create a copy of the contained value instead of moving out of src.

It is not a compile-time error if Src and Dst have different sizes, but it is highly encouraged to only invoke this function where Src and Dst have the same size. This function triggers undefined behavior if Dst is larger than Src.

Examples

use std::mem;

#[repr(packed)]
struct Foo {
    bar: u8,
}

let foo_array = [10u8];

unsafe {
    // Copy the data from 'foo_array' and treat it as a 'Foo'
    let mut foo_struct: Foo = mem::transmute_copy(&foo_array);
    assert_eq!(foo_struct.bar, 10);

    // Modify the copied data
    foo_struct.bar = 20;
    assert_eq!(foo_struct.bar, 20);
}

// The contents of 'foo_array' should not have changed
assert_eq!(foo_array, [10]);