I need help understanding the 'if let' statement and 'unwrapping'.

Hi Ryan,

Let's say that you have a couple of variables:

var a = 1
var b: Int?

In the variables above, you know that a has a value of 1, but let's pretend that you are not sure what the value of b is until after your user has done something in an application, say something like entering the number of siblings they have on a survey. We know it will be an integer, but we don't know what the actual value will be so we set it as an Optional.

When a variable is created without an initial value, say, when the survey is initially loaded with no default values, it is of type Optional. As you can see, a has a value of 1, while b has no assigned value, which defaults to nil. You can see this by printing the value of b:

println(b) // prints nil to the console

Now, using the variables above, I am using if let to safely get the value of b and assign it to c:

if let c = b {
    println("c has a value of \(c).")
} else {
    println("c is nil.")
}

This code will print "c is nil." because when we try to assign the value of b to a new constant named c on the first line, c is nil because b is nil. The way the line reads is this:

'If b has a value, assign it to a constant named c.'

Since b is nil, c cannot be assigned an integer value, and so c must also be nil, which executes the else statement.

Now, we will assign b an integer value:

var a = 1
var b: Int?

b = 2


if let c = b {
    println("c has a value of \(c).")
} else {
    println("c is nil.")
}

Since b now has a value of 2, when we try to assign c the value of b, which is now 2, c can successfully be created with a value of 2. "c has a value of 2." will be printed because c was successfully created as a constant with the valid integer value of b since b is no longer nil.

Just as before, if let basically says 'If b has a value, assign it to a constant named c.'

Hope this helps!