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SOLVED: Question: why name is a constant in my simple codes ?

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@boat  

Hello this is a pretty simple example code in the Playground , but at the line of name = name.uppercased(), Xcode warns me that error: cannot assign to value: 'name' is a 'let' constant

How come name is a constant ? I didn't declare it as a constant or a variable.


import Cocoa

func isUppercase(_ name: String) -> String {
   name = name.uppercased()
}

let result = isUppercase("Mary")

print (result)

   

Parameters are passed into functions as constant (i.e., let) values.

Side note: It's a bad idea to name that function isUppercase. Names like is______ in Swift (and most languages) would be expected to return a boolean indicating whether or not the parameter is uppercased. If you want to convert a String to uppercase, name it something else, like uppercase or toUppercase.

1      

@boat  

@roosterboy thank you as alway !

noted about the convention too.

Happy Lunar New Year and have a great year of tiger !

Boat

   

As @roosterboy say parameters passing in are none mutating however if you want them to be you need to add inout eg

func toUppercase(_ name: inout String) -> String {
   name.uppercased()
}

And when you call it you need a & in front of the variable that you want to mutate.

var mary = "Mary"
let result = toUppercase(&mary)
print(result) // MARY

More detail explanation What are inout parameters?

1      

@boat  

@NigelGee thanks for the extensive explanation !

Will look into it !

have a great day,

Boat

   

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