49

The documentation doesn't say how and the tutorial completely ignores for loops.

kmdreko
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mcandre
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    possible duplicate of [In rust, what is the idiomatic equivalent of Haskell's \[n..m\]?](http://stackoverflow.com/questions/15836096/in-rust-what-is-the-idiomatic-equivalent-of-haskells-n-m) – Mirzhan Irkegulov Dec 12 '14 at 08:02
  • Take a look at https://doc.rust-lang.org/std/ops/struct.RangeFrom.html – CrazyVideoGamer Mar 27 '21 at 13:31
  • This is an old and obsolete question, but you should have provided at least the link to the "tutorial" you were reading. We cannot just guess. – nbro Dec 31 '22 at 11:18

6 Answers6

39

As of 1.0, for loops work with values of types with the Iterator trait.

The book describes this technique in chapter 3.5 and chapter 13.2.

If you are interested in how for loops operate, see the described syntactic sugar in Module std::iter.

Example:

fn main() {
    let strs = ["red", "green", "blue"];

    for sptr in strs.iter() {
        println!("{}", sptr);
    }
}

(Playground)

If you just want to iterate over a range of numbers, as in C's for loops, you can create a numeric range with the a..b syntax:

for i in 0..3 {
    println!("{}", i);
}

If you need both, the index and the element from an array, the idiomatic way to get that is with the Iterator::enumerate method:

fn main() {
    let strs = ["red", "green", "blue"];

    for (i, s) in strs.iter().enumerate() {
        println!("String #{} is {}", i, s);
    }
}

Notes:

  • The loop items are borrowed references to the iteratee elements. In this case, the elements of strs have type &'static str - they are borrowed pointers to static strings. This means sptr has type &&'static str, so we dereference it as *sptr. An alternative form which I prefer is:

      for &s in strs.iter() {
      println!("{}", s);
  }
Peter Mortensen
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nejucomo
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12

for i in range(0, 100) is now deprecated in favour of for i in 0..100 (according to rustc 1.0.0-nightly.

Also worth noting is the compiler can't disambiguate when you use an identifier in the range (e.g., for i in 0..a), so you have to use for i in (0..a), but there's a pull request submitted to fix this.

Peter Mortensen
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confused00
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10

Actually, the Loops section of the tutorial does cover for loops:

When iterating over a vector, use for instead.

for elt in ["red", "green", "blue"] {
   std::io::println(elt);
}

But if you needed indices, you could do something like the following, using the uint::range function from the core library (or int::range or u8::range or u32::range or u64::range) and Rust's syntax for blocks:

range(0u, 64u, {|i| C[i] = A[i] + B[i]});

Rust used to support this equivalent syntax but it was later removed:

range(0u, 64u) {|i|
    C[i] = A[i] + B[i];
}
Warren P
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Lindsey Kuper
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    Thanks! I'm curious why Rust has two different syntaxes for function definitions and blocks. Seems like they could save the coder trouble by reusing `fn(args...)` instead of `|args|`. – mcandre Feb 14 '12 at 21:09
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    @mcandre Actually, the Ruby-style block syntax `{|args| body}` is used to denote a closure rather than just a function. It's also really convenient for simplifying usage of anonymous functions as you might see them used in Javascript, since any function that accepts a closure as its last argument (such as a callback) can be written _after_ the function call, as in Lindsey's third example above. Finally, even though Rust has a few different types of closures, Rust can infer the type of closure you want when using the block syntax. See also http://doc.rust-lang.org/doc/tutorial.html#closures – B. Striegel Feb 15 '12 at 15:18
  • Does this still work? I can't get range(n,n) {|i| ...} to compile. I get weird errors. – Warren P Apr 27 '13 at 18:24
  • As of rust 0.6, for int::range() |i| { C[i] = A[i] + B[i]; } should work. for is syntactic sugar which passes the closure in as an argument, iirc. –  Jun 05 '13 at 22:01
4
let range = (start..end).collect::<Vec<i32>>();
troglodytto
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0

Note that as of rustc 0.4 (Oct 2012), the alternate construction of

range(0u, 64u) {|i|
    C[i] = A[i] + B[i];
}

appears to not be supported any more.

Bob Hansen
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-4

Use int::range.

mcandre
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