How to use inout parameters?

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All function parameters in Swift are constants by default. That means you can not change the value of the parameters. Trying to change the value of a function parameter from within the body of that function results in a compile-time error. If you want a function to modify a parameter’s value then the parameter should be used as an inout parameter.

The inout parameters can be updated inside the function. The updated value replaces the original value outside the function.

func reverseName(_ name: inout String) {
    name = String(name.reversed()) + ".com"
}

var websiteName = "ishtiz"
print(websiteName)
reverseName(&websiteName)
print(websiteName)

// Output
// ishtiz
// zithsi.com

// The variable websiteName is changed.

There are some key restrictions that we want to remember when using inout parameters. You can’t use a constant(let) as the argument because constants can’t be changed. We can not set default values as inout parameters. The variadic parameters can’t be marked as inout as well. When you pass inout parameters as an argument you need to use an ampersand (&) directly before a variable. It will state that the function will modify the parameters.

Planning to apply for an iOS job? Check out this article to uplift your resume! Also helpful – SwiftUI and Swift Interview preparation. Happy job hunting!

Written by

ISHTIAK AHMED

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