Concatenate number with string in Swift

Question

I need to concatenate a String and Int as below:

let myVariable: Int = 8
return "first " + myVariable

But it does not compile, with the error:

Binary operator '+' cannot be applied to operands of type 'String' and 'Int'

What is the proper way to concatenate a String + Int?

Answer 1

If you want to put a number inside a string, you can just use String Interpolation:

return "first \(myVariable)"

Answer 2

You have TWO options;

return "first " + String(myVariable)

or

return "first \(myVariable)"

Answer 3

To add an Int to a String you can do:

return "first \(myVariable)"

Answer 4

If you're doing a lot of it, consider an operator to make it more readable:

func concat<T1, T2>(a: T1, b: T2) -> String {
    return "\(a)" + "\(b)"
}

let c = concat("Horse ", "cart") // "Horse cart"
let d = concat("Horse ", 17) // "Horse 17"
let e = concat(19.2345, " horses") // "19.2345 horses"
let f = concat([1, 2, 4], " horses") // "[1, 2, 4] horses"

operator infix +++ {}
@infix func +++ <T1, T2>(a: T1, b: T2) -> String {
    return concat(a, b)
}

let c1 = "Horse " +++ "cart"
let d1 = "Horse " +++ 17
let e1 = 19.2345 +++ " horses"
let f1 = [1, 2, 4] +++ " horses"

You can, of course, use any valid infix operator, not just +++.

Answer 5

Optional keyword would appear when you have marked variable as optional with ! during declaration.

To avoid Optional keyword in the print output, you have two options:

1. Mark the optional variable as non-optional. For this, you will have to give default value.
2. Use force unwrap (!) symbol, next to variable

In your case, this would work just fine

return "first \(myVariable!)"

Answer 6

var a = 2
var b = "Hello W"
print("\(a) " + b)

will print 2 Hello W

Answer 7

Here is documentation about String and characters

var variableString = "Horse"
variableString += " and carriage"
// variableString is now "Horse and carriage"