Function no_std_compat::ptr::read

1.0.0 (const: 1.71.0) · source ·
pub const unsafe fn read<T>(src: *const T) -> T
Expand 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:

  • src must be valid for reads.

  • src must be properly aligned. Use read_unaligned if this is not the case.

  • src must point to a properly initialized value of type T.

Note that even if T has size 0, the pointer must be non-null and 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 to be dropped.

use std::ptr;

let mut s = String::from("foo");
unsafe {
    // `s2` now points to the same underlying memory as `s`.
    let mut s2: String = ptr::read(&s);

    assert_eq!(s2, "foo");

    // Assigning to `s2` causes its original value to be dropped. Beyond
    // this point, `s` must no longer be used, as the underlying memory has
    // been freed.
    s2 = String::default();
    assert_eq!(s2, "");

    // Assigning to `s` would cause the old value to be dropped again,
    // resulting in undefined behavior.
    // s = String::from("bar"); // ERROR

    // `ptr::write` can be used to overwrite a value without dropping it.
    ptr::write(&mut s, String::from("bar"));
}

assert_eq!(s, "bar");