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iOS Swift Functions and Optionals Functions Syntax and Parameters

Romi Stepovich
Romi Stepovich
8,473 Points

I am getting this error message about my String `greeting` function

Here is the error message; Your greeting function printed "Hello (personNameString)" when called with "XYZZY" as the parameter. It should have printed "Hello XYZZY".

Here is my code. I know I am missing something but I am just not sure what it is...

func greeting(personName: String) { let personNameString = "Tom" println("Hello (personNameString)") }

3 Answers

Hi Romi, let's go step by step

at the begging of a challenge you have this piece of code

func greeting() {
    println("Hello")
}

First task is: Modify the function named greeting to accept a parameter. Name the parameter person which is of type String.

We need to add person: String in parentheses like this

func greeting(person: String) {
    println("Hello")
}

Second task is: Modify the println statement within the greeting function to use the variable person. For example, if you pass a person named "Tom" to the function then it should display: Hello Tom. (Hint: you must use string interpolation).

We need to add (person) to a string in println like this

func greeting(person: String) {
    println("Hello \(person)")
}

And finally third task is: Call the function greeting and pass it the string "Tom".

All we need to do is to call a function like this: greeting("Tom")

This is whole correct code

func greeting(person: String) {
    println("Hello \(person)")
}

greeting("Tom")
Romi Stepovich
Romi Stepovich
8,473 Points

Thank you, Petar! That makes far more sense. I tend to try and overthink code. That makes far more sense the way you laid it out for me. I appreciate your help!

Joseph F Fares Jr
Joseph F Fares Jr
517 Points

Thank you...Very clear explanation