using guard to check for nil without implicitly unwrapping

Question

I know there are some similar questions around, but I couldn't find one specific to my issue. I have a request where I want to check for the presence of the error key. it is not present everything is fine, if not I should handle the error. Currently, I have it implemented as follows:

if let error = json["error"] {
    // handle error
}
else {
    // handle success
}

I would like to use a guard statement here to have the success case unindented. The only way I came up with is

guard json["error"] == nil else {
    let error = json["error"]!
    // handle error
} 

// handle success

but that seems wrong to me with the !. Are there any other approaches to this?

Answer 1

In your guard code you would have to have a return statement in the else block. Like this...

guard json["error"] == nil else {
    let error = json["error"]!
    // handle error
    return
}

// handle success

But you are correct. Having to force unwrap the error is not ideal.

So in this case. I think guard is the wrong solution. Instead use if but return from the conditional block. This removes the need for using an else block.

if let error = json["error"] {
    print(error)
    // handle error
    return
}

// handle success...
// no need for else block. Just return from the if in the error case.

The difference between guard let and if let is where the unwrapped optional is scoped.

With guard it is scoped outside the block with if it is scoped inside the block.