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Update the ways to get a pointer from a box
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Show how to get a pointer without destroying the box.
Use `boxed::into_raw` instead of `mem::transmute`.
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gkoz committed Mar 12, 2015
1 parent 49f7550 commit 5d9ed0b
Showing 1 changed file with 16 additions and 13 deletions.
29 changes: 16 additions & 13 deletions src/libcore/ptr.rs
Expand Up @@ -33,31 +33,34 @@
//! let my_speed_ptr: *mut i32 = &mut my_speed;
//! ```
//!
//! To get a pointer to a boxed value, dereference the box:
//!
//! ```
//! let my_num: Box<i32> = Box::new(10);
//! let my_num_ptr: *const i32 = &*my_num;
//! let mut my_speed: Box<i32> = Box::new(88);
//! let my_speed_ptr: *mut i32 = &mut *my_speed;
//! ```
//!
//! This does not take ownership of the original allocation
//! and requires no resource management later,
//! but you must not use the pointer after its lifetime.
//!
//! ## 2. Transmute an owned box (`Box<T>`).
//! ## 2. Consume a box (`Box<T>`).
//!
//! The `transmute` function takes, by value, whatever it's given
//! and returns it as whatever type is requested, as long as the
//! types are the same size. Because `Box<T>` and `*mut T` have the same
//! representation they can be trivially,
//! though unsafely, transformed from one type to the other.
//! The `into_raw` function consumes a box and returns
//! the raw pointer. It doesn't destroy `T` or deallocate any memory.
//!
//! ```
//! use std::mem;
//! use std::boxed;
//!
//! unsafe {
//! let my_num: Box<i32> = Box::new(10);
//! let my_num: *const i32 = mem::transmute(my_num);
//! let my_speed: Box<i32> = Box::new(88);
//! let my_speed: *mut i32 = mem::transmute(my_speed);
//! let my_speed: *mut i32 = boxed::into_raw(my_speed);
//!
//! // By taking ownership of the original `Box<T>` though
//! // we are obligated to transmute it back later to be destroyed.
//! drop(mem::transmute::<_, Box<i32>>(my_speed));
//! drop(mem::transmute::<_, Box<i32>>(my_num));
//! // we are obligated to put it together later to be destroyed.
//! drop(Box::from_raw(my_speed));
//! }
//! ```
//!
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