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# Why does assigning the return value of the function to a constant created in the if condition solve the error message?

So given a correctly written isDivisible function, if we do:

<pre><code> if isDivisible(num1, num2) { \\ do something } else { \\ do other thing } </code></pre>

xCode gives an error that the optional type bool? cannot be used as a boolean. If we then instead do:

let result = isDivisible(num1, num2)

if result { \\ do something } else { \\ do other thing }

xCode gives the same error as above, saying that optional type bool? can't be used as a boolean. But how come the following code works without erroring?

if let result = isDivisible(num1, num2) { \\ do something } else { \\ do other thing }

Why does assigning the return value of the isDivisible function to a constant declare in the if condition pass the error check when the other two types don't?

the function isDivisible returns an optional `isDivisible(num1, num2) -> Bool?`, so if you want to get it's value you have to unwrap it (using the !).

`let result = isDivisible(num1, num2)` is an optional of type Bool? and not a Bool, to use it's value you need to unwrap it first.

```// result is an optional of type Bool?
let result = isDivisible(num1, num2)

// result! is the value of the optional might be of type Bool or nil
if result! {
// Do something
} else {
// Do something else
}
```

the `if let` it's an optional binding that makes it easier to test if an optional has a value.

```if let result = isDivisible(num1, num2) {
// Do Something
} else {
// Do something else
}
```

In this conditional, the existence of a value returned by the function is tested first. If the returned value is nil, the else statement is executed. If is not nil, the optional returned by the function is unwrapped and assigned to the constant result.