str::contains doesn't work when the supplied input is a string reference [duplicate]

Question

use std::collections::HashSet; fn character_count(needles: &HashSet<char>, needle_type: &str, input: &str) -> i32 { for needle in needles { let mut needle_custom; if needle_type == "double" { needle_custom = needle.to_string() + &needle.to_string(); } else { needle_custom = needle.to_string() + &needle.to_string() + &needle.to_string(); } if input.contains(needle_custom) { println!("found needle {:?}", needle_custom); } } return 1; }

error[E0277]: expected a `std::ops::FnMut<(char,)>` closure, found `std::string::String` --> src/lib.rs:13:18 | 13 | if input.contains(needle_custom) { | ^^^^^^^^ expected an `FnMut<(char,)>` closure, found `std::string::String` | = help: the trait `std::ops::FnMut<(char,)>` is not implemented for `std::string::String` = note: required because of the requirements on the impl of `std::str::pattern::Pattern<'_>` for `std::string::String`

The code works if I replace needle_custom with "test".

Its kind of a bummer that the error message mentions char because it makes it harder to diagnose as a common &str vs String issue — turbulencetoo
– turbulencetoo, Commented Feb 22, 2019 at 21:49
In the future, you are encouraged to search for your exact error message. In many cases, this will directly lead you to existing questions and answers. — Shepmaster
– Shepmaster, Commented Feb 24, 2019 at 19:35

Shepmaster · Accepted Answer · 2019-02-24 19:33:56Z

The contains method will accept an &str or a char but not a String.

The contains method is declared as

pub fn contains<'a, P>(&'a self, pat: P) -> bool where P: Pattern<'a>,

And if you look at the implementors of Pattern, you'll see its implemented for char and for &str.

This means that you need to pass a &str to contains instead of your owned String. Because &String coerces to &str, this is an easy change:

-- if input.contains(needle_custom) { ++ if input.contains(&needle_custom) {

Here is your code with this small change working in the Playground.

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