Auto merge of #26664 - steveklabnik:gh26571, r=alexcrichton

This was pretty misleading, so let's improve.

Fixes #26571
This commit is contained in:
bors 2015-06-30 12:54:16 +00:00
commit 0b703787ab

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@ -365,11 +365,48 @@ pub fn replace<T>(dest: &mut T, mut src: T) -> T {
/// Disposes of a value.
///
/// This function can be used to destroy any value by allowing `drop` to take ownership of its
/// argument.
/// While this does call the argument's implementation of `Drop`, it will not
/// release any borrows, as borrows are based on lexical scope.
///
/// # Examples
///
/// Basic usage:
///
/// ```
/// let v = vec![1, 2, 3];
///
/// drop(v); // explicitly drop the vector
/// ```
///
/// Borrows are based on lexical scope, so this produces an error:
///
/// ```ignore
/// let mut v = vec![1, 2, 3];
/// let x = &v[0];
///
/// drop(x); // explicitly drop the reference, but the borrow still exists
///
/// v.push(4); // error: cannot borrow `v` as mutable because it is also
/// // borrowed as immutable
/// ```
///
/// An inner scope is needed to fix this:
///
/// ```
/// let mut v = vec![1, 2, 3];
///
/// {
/// let x = &v[0];
///
/// drop(x); // this is now redundant, as `x` is going out of scope anyway
/// }
///
/// v.push(4); // no problems
/// ```
///
/// Since `RefCell` enforces the borrow rules at runtime, `drop()` can
/// seemingly release a borrow of one:
///
/// ```
/// use std::cell::RefCell;
///