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auto merge of #15899 : aochagavia/rust/guide, r=kballard

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The removed code caused confusion because it is not clear that the type of `y` is actually `()`
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bors 2014-07-23 06:31:11 +00:00
commit 83a8a56473
1 changed files with 4 additions and 9 deletions
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src/doc/guide.md
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@ -624,15 +624,10 @@ let x = (let y = 5i); // found `let` in ident position
The compiler is telling us here that it was expecting to see the beginning of
an expression, and a `let` can only begin a statement, not an expression.
However, assigning to a variable binding is an expression:
```{rust}
let x;
let y = x = 5i;
```
In this case, we have an assignment expression (`x = 5`) whose value is
being used as part of a `let` declaration statement (`let y = ...`).
Note that assigning to an already-bound variable (e.g. `y = 5i`) is still an
expression, although its value is not particularly useful. Unlike C, where an
assignment evaluates to the assigned value (e.g. `5i` in the previous example),
in Rust the value of an assignment is the unit type `()` (which we'll cover later).
The second kind of statement in Rust is the **expression statement**. Its
purpose is to turn any expression into a statement. In practical terms, Rust's