Default parameter is not leveraged when the variable that is assigned the function with the default argument is called

[andrewkykoo]

Description

1. There is a function that takes a default value for a parameter
2. Let's define a variable as a function type by assigning the function above to the variable
3. When the variable is called without a parameter, it does not leverage the default parameter

Steps to reproduce

func showNumber(num: Int = 10) {
    print(num)
}

var showNum = showNumber

showNum()    // throws an error ("Missing argument for parameter #1 in call")

// modified to take an optional Int? argument
func showNumber(num: Int? = 10) {
    if let num = num {
        print(num)
    } else {
        print("num is nil")
    }

...

showNum()    // still throws the same error

Expected behavior

• Expected showNum() to print 10

Environment

• Swift compiler version info : swift-driver version: 1.75.2 Apple Swift version 5.8 (swiftlang-5.8.0.124.2 clang-1403.0.22.11.100) Target: arm64-apple-macosx13.0
• Xcode version info : Xcode 14.3 Build version 14E222b
• Deployment target: : macOS

[xedin]

This is expected behavior, in swift values of function type do not have default values applied, showNum has type (Int?) -> Void

[andrewkykoo]

This is expected behavior, in swift values of function type do not have default values applied, showNum has type (Int?) -> Void

Since this is an expected behavior, do you suggest not assigning functions that take default parameters to variables of function type?

[xedin]

They can be assigned, just don’t expect that they can be called without providing arguments.

[andrewkykoo]

Okay. I will keep in mind that variables of function type that are assigned functions with default parameters have to be called with arguments. In other words, they don't take default arguments.

At one point, I was wondering if the type could look like this, but this is simply not a correct type defining syntax for variables of function type.

var showNum: (Int? = 10) -> Void

Thanks for clarifying this!

[AnthonyLatsis]

There is no reason not to pass around functions with defaulted parameters as values, but as Pavel mentioned, function types do not carry info about default values or arguments labels, so as soon as you turn a function declaration reference into a function value, all that info is lost. All that is left is the function type, with no association to a function declaration. For example, showNum(num: 10) is also invalid, you can only do showNum(10).

Moreover, the type-checker does not attempt to guess evaluation results (e.g. that showNum in showNum() references showNumber). Roughly speaking, all it cares about is the kind of expression (showNum is a reference to a variable) and its type (showNum is of type (Int) -> Void).