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snippetrustMajor

How do I sum a vector using fold?

Submitted by: @import:stackoverflow-api··
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howusingsumvectorfold

Problem

This Rust tutorial explains the fold() mechanism well, and this example code:

let sum = (1..4).fold(0, |sum, x| sum + x);


works as expected.

I'd like to run it on a vector, so based on that example, first I wrote this:

let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0, |sum, val| sum += val);


which threw an error:

error: binary assignment operation `+=` cannot be applied to types `_` and `&u32` [E0368]
let sum = ratings.values().fold(0, |sum, val| sum += val);
                                              ^~~~~~~~~~


I guessed this might be a reference-related error for some reason, so I changed that to fold(0, |sum, &val| sum += val), which resulted in

error: mismatched types:
expected `u32`,
   found `()`


Hm, maybe something's wrong with the closure? Using {sum += x; sum }, I got

binary assignment operation `+=` cannot be applied to types `_` and `&u32`


again.

After further trial and error, adding mut to sum worked:

let sum = vec![1,2,3,4,5,6].iter().fold(0, |mut sum, &x| {sum += x; sum});


Could someone explain the reason why fold() for vectors differs so much from the tutorial? Or is there a better way to handle this?

For reference, I'm using Rust beta, 2015-04-02.

Solution

Since Rust 1.11, you can sum the iterator directly, skipping fold:

let sum: u32 = vec![1, 2, 3, 4, 5, 6].iter().sum();


You've already figured out that += is the problem, but I'd like to provide some more exposition.

In your case, the arguments provided to the fold closure are _ and &u32. The first type is an not-yet-specified integer. If you change your fold call to fold(0u32, |sum, val| sum += val), you'll get a slightly different message:

let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0u32, |sum, val| sum += val);


error[E0308]: mismatched types
|
2 | let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0u32, |sum, val| sum += val);
| ^^^ expected u32, found &{integer}
|
= note: expected type
u32
= note: found type
&{integer}


The result value of the binary assignment operation += is (), the unit type. This explains the error message when you changed to fold(0, |sum, &val| sum += val):

let mut a = 1;
let what_am_i = a += 1;
println!("{:?}", what_am_i); // => ()


If you change to fold(0, |sum, &val| {sum += val ; sum}), you then get an understandable error about immutable variables:

let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0, |sum, &val| {sum += val; sum});


error[E0384]: re-assignment of immutable variable sum
--> src/main.rs:2:66
|
2 | let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0, |sum, &val| {sum += val; sum});
| --- ^^^^^^^^^^ re-assignment of immutable variable
| |
| first assignment to
sum


From here, you could mark sum as mutable, but the correct solution is to simply fold with sum + val, as you discovered.

Code Snippets

let sum: u32 = vec![1, 2, 3, 4, 5, 6].iter().sum();
let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0u32, |sum, val| sum += val);
let mut a = 1;
let what_am_i = a += 1;
println!("{:?}", what_am_i); // => ()
let sum: u32 = vec![1,2,3,4,5,6].iter().fold(0, |sum, &val| {sum += val; sum});

Context

Stack Overflow Q#29548819, score: 72

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