Understanding Rust Error E0627: Yield Expression Outside of Generator Literal

Introduction to Rust Error E0627

Rust Error E0627 occurs when the 'yield' keyword is used outside of a generator literal. In this article, we will delve into the error in more detail, explaining its cause and providing a solution to fix it.

Erroneous Code Example

#![feature(generators, generator_trait)]

fn fake_generator() -> &'static str {
    yield 1;
    return "foo"
}

fn main() {
    let mut generator = fake_generator;
}

The error in the code example above occurs because the 'yield' keyword is used within the 'fake_generator' function, which is not a generator literal. The 'yield' keyword is reserved for use within generator literals only.

Fixing Rust Error E0627

To fix Rust Error E0627, ensure that 'yield' expressions are used within generator literals. Refactoring the above erroneous code to correctly define the generator literal would resolve the error.

Corrected Code Example

#![feature(generators, generator_trait)]

fn main() {
    let mut generator = || {
        yield 1;
        return "foo"
    };
}

In this revised code example, the 'yield' expression is correctly used within a generator literal. As a result, the Rust Error E0627 is no longer an issue.

Conclusion

Rust Error E0627 arises when a 'yield' expression is employed outside of a generator literal. To resolve this error, ensure that the 'yield' expression is placed within the correct generator literal context. Following this article's guidance to refactor problematic code will eliminate Rust Error E0627 and ensure smooth program execution.