pub unsafe fn write<T, const host: bool = true>(dst: *mut T, src: T)
Expand description
Overwrites a memory location with the given value without reading or dropping the old value.
write
does not drop the contents of dst
. This is safe, but it could leak
allocations or resources, so care should be taken not to overwrite an object
that should be dropped.
Additionally, it does not drop src
. Semantically, src
is moved into the
location pointed to by dst
.
This is appropriate for initializing uninitialized memory, or overwriting
memory that has previously been read
from.
Safety
Behavior is undefined if any of the following conditions are violated:
-
dst
must be valid for writes. -
dst
must be properly aligned. Usewrite_unaligned
if this is not the case.
Note that even if T
has size 0
, the pointer must be non-null and properly aligned.
Examples
Basic usage:
let mut x = 0;
let y = &mut x as *mut i32;
let z = 12;
unsafe {
std::ptr::write(y, z);
assert_eq!(std::ptr::read(y), 12);
}
RunManually implement mem::swap
:
use std::ptr;
fn swap<T>(a: &mut T, b: &mut T) {
unsafe {
// Create a bitwise copy of the value at `a` in `tmp`.
let tmp = ptr::read(a);
// Exiting at this point (either by explicitly returning or by
// calling a function which panics) would cause the value in `tmp` to
// be dropped while the same value is still referenced by `a`. This
// could trigger undefined behavior if `T` is not `Copy`.
// Create a bitwise copy of the value at `b` in `a`.
// This is safe because mutable references cannot alias.
ptr::copy_nonoverlapping(b, a, 1);
// As above, exiting here could trigger undefined behavior because
// the same value is referenced by `a` and `b`.
// Move `tmp` into `b`.
ptr::write(b, tmp);
// `tmp` has been moved (`write` takes ownership of its second argument),
// so nothing is dropped implicitly here.
}
}
let mut foo = "foo".to_owned();
let mut bar = "bar".to_owned();
swap(&mut foo, &mut bar);
assert_eq!(foo, "bar");
assert_eq!(bar, "foo");
Run