Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
548 views
in Technique[技术] by (71.8m points)

recursion - Recursive function if statement mismatched types in Rust

fn recursive_binary_search<T: Ord>(list: &mut [T], target: T) -> bool {
    if list.len() < 1 {
        return false;
    }
    let guess = list.len() / 2;
    if target == list[guess] {
        return true;
    } else if list[guess] > target {
        return recursive_binary_search(&mut list[0..guess], target);
    } else if list[guess] < target {
        return recursive_binary_search(&mut list[guess..list.len()], target);
    }
}

the compiler throws an error on if target == list[guess] saying

src/main.rs:33:5: 39:6 error: mismatched types [E0308]
src/main.rs:33     if target == list[guess] {
                   ^
src/main.rs:33:5: 39:6 help: run `rustc --explain E0308` to see a detailed explanation
src/main.rs:33:5: 39:6 note: expected type `bool`
src/main.rs:33:5: 39:6 note:    found type `()`
error: aborting due to previous error

I can't figure out how to rewrite this function to satisfy the type checker. I assume it is because I have the return type set to bool and there is a return function call?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The issue here is that Rust evaluates the if/else if/else if block as the return value because it lacks an else clause, and statements which don't evaluate to any value have the type (). Incidentally, the code you've presented does exhaustively cover all possibilities (the item at the current index of the slice is either equal to, less than, or greater than the target), but the compiler doesn't know that unless you give it an else clause at the end:

fn recursive_binary_search<T: Ord + Eq>(list: &[T], target: T) -> bool {
    if list.len() < 1 {
        return false;
    }
    let guess = list.len() / 2;
    if target == list[guess] {
        return true;
    } else if list[guess] > target {
        return recursive_binary_search(&list[0..guess], target);
    } else {
        return recursive_binary_search(&list[guess..list.len()], target);
    }
}

PS: This function doesn't require mutable references, so I'd recommend using regular references as in my code above.

EDIT: For posterity, here's the same code w/o explicit returns:

fn recursive_binary_search<T: Ord>(list: &[T], target: T) -> bool {
    if list.len() < 1 {
        return false;
    }
    let guess = list.len() / 2;
    if target == list[guess] {
        true
    } else if list[guess] > target {
        recursive_binary_search(&list[0..guess], target)
    } else {
        recursive_binary_search(&list[guess..list.len()], target)
    }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...