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

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	@boat Feb '22 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)` 2
	@roosterboy HWS+ Feb '22 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`. 3
	@boat Feb '22 `@roosterboy` thank you as alway ! noted about the convention too. Happy Lunar New Year and have a great year of tiger ! Boat 2
	@NigelGee HWS+ Feb '22 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? 3
	@boat Feb '22 `@NigelGee` thanks for the extensive explanation ! Will look into it ! have a great day, Boat 2

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