pub const unsafe fn read<T>(src: *const T) -> TExpand description
Reads the value from src without moving it. This leaves the
memory in src unchanged.
§Safety
Behavior is undefined if any of the following conditions are violated:
-
srcmust be valid for reads. -
srcmust be properly aligned. Useread_unalignedif this is not the case. -
srcmust point to a properly initialized value of typeT.
Note that even if T has size 0, the pointer must be properly aligned.
§Examples
Basic usage:
let x = 12;
let y = &x as *const i32;
unsafe {
assert_eq!(std::ptr::read(y), 12);
}Manually 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");§Ownership of the Returned Value
read creates a bitwise copy of T, regardless of whether T is Copy.
If T is not Copy, using both the returned value and the value at
*src can violate memory safety. Note that assigning to *src counts as a
use because it will attempt to drop the value at *src.
write() can be used to overwrite data without causing it