Concatenate number with string in Swift

StringNsstringSwift

String Problem Overview


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?

String Solutions


Solution 1 - String

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

return "first \(myVariable)"

Solution 2 - String

You have TWO options;

return "first " + String(myVariable)

or

return "first \(myVariable)"

Solution 3 - String

To add an Int to a String you can do:

return "first \(myVariable)"

Solution 4 - String

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 +++.

Solution 5 - String

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!)"

Solution 6 - String

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

will print 2 Hello W

Solution 7 - String

Here is documentation about String and characters

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

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionBegumView Question on Stackoverflow
Solution 1 - StringjtbandesView Answer on Stackoverflow
Solution 2 - StringH. MahidaView Answer on Stackoverflow
Solution 3 - StringCW0007007View Answer on Stackoverflow
Solution 4 - StringGrimxnView Answer on Stackoverflow
Solution 5 - StringSategroupView Answer on Stackoverflow
Solution 6 - Stringvijay alapatiView Answer on Stackoverflow
Solution 7 - StringTomasz SzulcView Answer on Stackoverflow