How to make a reverse ordered for loop?

Question

Editor's note: This question was asked before Rust 1.0 was released and the .. "range" operator was introduced. The question's code no longer represents the current style, but some answers below uses code that will work in Rust 1.0 and onwards.

I was playing on the Rust by Example website and wanted to print out fizzbuzz in reverse. Here is what I tried:

fn main() {
    // `n` will take the values: 1, 2, ..., 100 in each iteration
    for n in std::iter::range_step(100u, 0, -1) {
        if n % 15 == 0 {
            println!("fizzbuzz");
        } else if n % 3 == 0 {
            println!("fizz");
        } else if n % 5 == 0 {
            println!("buzz");
        } else {
            println!("{}", n);
        }
    }
}

There were no compilation errors, but it did not print out anything. How do I iterate from 100 to 1?

Shepmaster · Accepted Answer · 2017-11-13 23:52:04Z

344

A forward loop is like this:

for x in 0..100 {
    println!("{}", x);
}

And a reverse loop is done by calling Iterator::rev to reverse the order:

for x in (0..100).rev() {
    println!("{}", x);
}

edited Nov 13, 2017 at 23:52

Shepmaster

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answered Jul 26, 2015 at 22:28

Luxspes

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2 Comments

Bill Yan Over a year ago

in rust, 0..10 means from 0 to 9, 10 is not included. so does (0..10).rev() mean "10 to 1" or "9 to 0"?

jtst Over a year ago

The parentheses help clear this up. The range will be evaluated first (0 to 9), which is then reversed (9 to 0).

Shepmaster · Accepted Answer · 2018-01-10 03:42:46Z

77

Editor's note: This question refers to parts of Rust that predate Rust 1.0. Look at other answers for up to date code.

Your code doesn't work because a uint of value -1 is equal the maximum value of uint. The range_step iterator stops immediately upon detecting overflow. Using an int fixes this.

std::iter::range_step(100i, 0, -1)

You can also reverse an iterator with rev().

for n in range(0u, 100).rev() { ... }

Though note that this will be from 99->0, rather than 100->1.

edited Jan 10, 2018 at 3:42

Shepmaster

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answered Aug 6, 2014 at 21:05

A.B.

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7 Comments

Tybalt Over a year ago

So why doesn't the compiler complain about mixing the types?

Matthieu M. Over a year ago

@user1452670: You are not mixing types, 100u is of type uint, 0 and -1 are untyped integrals for which the compiler will infer a type. Here, range_step requires uniform arguments and receives (uint, <untyped>, <untyped>) so the compiler deduces they are all uint. If you had specified 0i, or -1i then you would have had a mismatch.

sellibitze Over a year ago

@MatthieuM.: To be fair, the value of -1 is known to the compiler and "inferring" its type to be uint could be made to produce a warning because it's not representable. The question is, how often this is an error or not.

Matthieu M. Over a year ago

@sellibitze: Certainly, a lint that would diagnose out-of-range assignments would be very welcome.

unegare Over a year ago

The first one is obsolete and the second one is unstable. As things stand in 2022. %)

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