Mar 17, 2024 / PROGRAMMING, CODING, RUST, KOTLIN

The pitfall of implicit returns

Implicit returns are a feature in some languages. They have recently bitten me, so here’s my opinion.

Statements, expressions, and returns

Before diving into implicit returns, we must explain two programming concepts influencing them. A lot of literature is available on the subject, so I’ll paraphrase one of the existing definitions:

An expression usually refers to a piece of code that can be evaluated to a value. In most programming languages, there are typically three different types of expressions: arithmetic, character, and logical.

A statement refers to a piece of code that executes a specific instruction or tells the computer to complete a task.

— Expression vs. Statement

Here’s a Kotlin snippet:

val y = 10                (1)
val x = 2                 (1)

x + y                     (2)

println(x)                (1)

1	Statement, executes the assignment "task"
2	Expression, evaluates to a value, e.g., 12

Functions may or may not return a value. When they do, they use the return keyword in most programming languages. It’s a statement that needs an expression.

In Kotlin, it translates to the following:

fun hello(who: String): String {
    return "Hello $who"
}

In this regard, Kotlin is similar to other programming languages with C-like syntax.

Implicit returns

A couple of programming languages add the idea of implicit returns: Kotlin, Rust, Scala, and Ruby are the ones I know about; each has different quirks.

I’m most familiar with Kotlin: you can omit the return keyword when you switch the syntax from a block body to an expression body. With the latter, you can rewrite the above code as the following:

fun hello(who: String): String = "Hello $who"

Rust also allows implicit returns with a slightly different syntax.

fn hello(who: &str) -> String {
    return "Hello ".to_owned() + who;          (1)
}

fn hello_implicit(who: &str) -> String {
    "Hello ".to_owned() + who                  (2)
}

1	Explicit return
2	Transform the statement in expression by removing the trailing semicolon - implicit return

Let’s continue with Kotlin. The expression doesn’t need to be a one-liner. You can use more complex expressions:

fun hello(who: String?): String =
    if (who == null) "Hello world"
	else "Hello $who"

The pitfall

I was writing code lately, and I produced something akin to this snippet:

enum class Constant {
    Foo, Bar, Baz
}


fun oops(constant: Constant): String = when (constant) {
    Constant.Foo -> "Foo"
    else -> {
        if (constant == Constant.Bar) "Bar"
        "Baz"
    }
}

Can you spot the bug? Let’s use the function to make it clear:

fun main() {
	println(oops(Constant.Foo))
	println(oops(Constant.Bar))
	println(oops(Constant.Baz))
}

The results are:

Foo
Baz
Baz

The explanation is relatively straightforward. if (constant == Constant.Bar) "Bar" does nothing. The following line evaluates to "Bar"; it implicitly returns the expression. To fix the bug, we need to add an else to transform the block into an expression:

if (constant == Constant.Bar) "Bar"
else "Baz"

Note that for simpler expressions, the compiler is smart enough to abort with an error:

fun oops(constant: Constant): String =
    if (constant == Constant.Bar) "Bar"      (1)
    "Baz"

1	`'if' must have both main and 'else' branches if used as an expression`

Conclusion

Implicit return is a powerful syntactic sugar that allows for more concise code. However, concise code doesn’t necessarily imply being better code. I firmly believe that explicit code is more maintainable in most situations.

In this case, I was tricked by my code! Beware of implicit returns.

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