Function std::mem::transmute_copy

pub unsafe fn transmute_copy<T, U>(src: &T) -> U

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

This function will unsafely assume the pointer src is valid for size_of::<U> bytes by transmuting &T to &U and then reading the &U (except that this is done in a way that is correct even when &U makes stricter alignment requirements than &T). 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 T and U have different sizes, but it is highly encouraged to only invoke this function where T and U have the same size. This function triggers undefined behavior if U is larger than T.

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]);