Concatenate number with string in Swift

55,906

Solution 1

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

return "first \(myVariable)"

Solution 2

You have TWO options;

return "first " + String(myVariable)

or

return "first \(myVariable)"

Solution 3

To add an Int to a String you can do:

return "first \(myVariable)"

Solution 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 +++.

Solution 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!)"

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Begum
Author by

Begum

Updated on July 16, 2022

Comments

  • Begum
    Begum almost 2 years

    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?

  • CW0007007
    CW0007007 almost 10 years
    The question is about concatenating it with an int.
  • SHA2NK
    SHA2NK almost 7 years
    It give back "Optional(value)" when used like above.