Type mismatch when returning from inside if statement [duplicate]

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The following code fails to compile:

fn testing() -> i32 {
let x = 5;
if x == 5 {
5
}
6
}


with this error:

error E0308: mismatched types. (expected (), found integral variable)


If I put an explicit return in front of the 5, or if I put the 6 inside an else block, everything works fine. What exactly is Rust complaining about?

return

5

6

else

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2 Answers
2

In Rust, nearly everything is an expression, including if-else. if generates a value, but there is no value for the expression if the expression fails (i.e. x != 5). When putting the 6 into the else block, the if-else statement returns an integral (which will be an i32 due to the return type). You can suppress the statement's value with a semicolon.

if

else

if

x != 5

6

if

else

i32

return 5 (note, that there is no semicolon) is also a statement, which results in (). Now the if expression always returns (), which is fine.

return 5

()

if

()

Idiomatically you should prefer the else variant, so you can omit the return:

else

return

fn testing() -> i32 {
let x = 5;
if x == 5 {
5
} else {
6
}
}


if-else is an assignable expression that has a type. Omitting else {} is equivalent to if { ... } else { () }. If you assign that value to a variable:

if-else

else {}

if { ... } else { () }

// would not compile
let result = if x == 5 {
5
};


it is equivalent to writing:

let result = if x == 5 {
5
} else {
()
};


Making the type of result either integer or unit, which is invalid.

result