Swift - Resolving a math operation in a string

just a short question. In Swift it is possible to solve the following code:

var a: String;
a = "\(3*3)";

The arithmetic operation in the string will be solved. But i can´t figure out, why this following variation doesn´t work.

var a: String; 
var b: String;

b = "3*3";
a = "\(b)";

In this case the arithmetic operation in var a will not be resolved. Any ideas why and how i can this get to work. Some things would be much more easier if this would work. Thanks for your answers.

Solution 1:

In the second case, you are interpolating a string, not an arithmetic expression. In your example, it's a string you chose at compile time, but in general it might be a string from the user, or loaded from a file or over the web. In other words, at runtime b could contain some arbitrary string. The compiler isn't available at runtime to parse an arbitrary string as arithmetic.

If you want to evaluate an arbitrary string as an arithmetic formula at runtime, you can use NSExpression. Here's a very simple example:

let expn = NSExpression(format:"3+3")
println(expn.expressionValueWithObject(nil, context: nil))
// output: 6

You can also use a third-party library like DDMathParser.

Solution 2:

Swift 4.2

let  expn = "3+3"
print(expn.expressionValue(with: nil, context: nil))

But I also have a solution thats not the most effective way but could be used in some cases if your sure it's only "y+x" and not longer string.

var yNumber: Int!
var xNumber: Int!

    let expn: String? = "3+3"
    // Here we take to first value in the expn String.
    if let firstNumber = expo?.prefix(1), let myInt = Int(firstNumber){
        // This will print (Int : 3)
        print("Int : \(myInt)")
        // I set the value to yNumber
        yNumber = myInt
    }
    // Here we take the last value in the expn string
    if let lastNumber = optionalString?.suffix(1), let myInt = Int(lastNumber){

        // This will print (Int : 3)
        print("Int : \(myInt)")
        // I set the value to xNumber
        xNumber = myInt
    }

  // Now you can take the two numbers and add
  print(yNumber + xNumber)
  // will print (6)

I can't recommend this but it works in some cases

Solution 3:

This won't be solved because this is not an arithmetic operation, this is a string:

"3*3"

the same as this

"String"

Everything you put in " it's a string.

The second example lets you construct a new String value from a mix of constants, variables, literals, and expressions:

"\(3*3)"

this is possible because of string interpolation \()

You inserted a string expression which swing convert and create expected result.