How to swizzle Swift.print(items:separator:terminator)

Question

I am looking for ways to swizzle Swift.print function. Overriding it is not an option, as it may get bypasses if you use Swift.print(:)

The selector does not recognise the identifier:

@objc class func printSwizzle() {
    guard let instance = class_getInstanceMethod(self, #selector(print(separator:terminator:))),
    let swizzleInstance = class_getInstanceMethod(self, #selector(swizzlePrint(separator:terminator:))) else { return }
    method_exchangeImplementations(instance, swizzleInstance)
}

Is that even possible? As swizzling is an obj-c runtime feature.

For a function to be swizzled, it has to be declared as `@dynamic` and marked with `@objc`. [Swift.print](https://github.com/apple/swift/blob/832307c30a600c7503d296139c58931c64f9c064/stdlib/public/core/Print.swift#L52) doesn't check any of those conditions. — ielyamani, Jan 01 '19 at 08:12
To add to what `Carpsen90` said, swizzling only works because in ObjC land, there is one central place that all method calls jump through, which is the ObjC runtime. That's great because it can let you do crazy stuff like intercepting and forwarding messages, swizzling method implementations, etc. On the down side, it can hurt performance pretty drastically. Swift function calls can be much more aggressively optimized (e.g. for small enough functions, calls can be replaced with an inline "copy", saving cost of a function call) — Alexander, Jan 01 '19 at 08:28
`Swift.print` is a global function, which doesn't belong to any class or object. It has no relation to the ObjC runtime, so you can't use the ObjC runtime to try to modify it — Alexander, Jan 01 '19 at 08:29
Thank you @matt , I was after swizzling the print function but as we all know and after everyone commented, this is not possible. All the answers are great for the community to learn. This is what I do now anyway, but wanted to avoid it. Cheers — Tal Zion, Jan 01 '19 at 19:32

Sulthan · Accepted Answer · 2019-01-01T19:35:05.023

Method swizzling is an Objective-C feature that enables you to exchange implementation of a method at runtime. For that, you need an @objc object that inherits from NSObject. And you need a method.

Swift.print is not a method. It's a function declared in the Swift module. We can say it's global but it's not really global. It's defined inside module Swift which is imported automatically to every Swift code, therefore you can use it without the Swift. prefix.

In summary, there is no way to swizzle Swift.print.

What you can do is to hide that function using your own implementation, that is, if you declare a function with the same name in your own module, then when print is used, the compiler will prefer your function instead because functions in the current module are preferred over functions in other modules (including Swift. module).

public func print(_ items: Any..., separator: String = " ", terminator: String = "\n") {
    let output = items.map { "\($0)" }.joined(separator: separator)
    Swift.print(output, terminator: terminator)
}

You can any logic you want in there.

It's actually very common to use this to remove logging from production, e.g.:

#if !DEBUG

func print(_ items: Any..., separator: String = " ", terminator: String = "\n") {}
func debugPrint(_ items: Any..., separator: String = " ", terminator: String = "\n") {}

#endif

See Remove println() for release version iOS Swift for more details.

Essentially, you could hide the whole Swift module by redeclaring it inside your module, e.g. as an enum, therefore disabling calls to Swift.print:

enum Swift {
    public static func print(_ items: Any..., separator: String = " ", terminator: String = " ") {
        // do something
    }
}

However, I would generally advise against this because it will be hard to solve any naming conflicts with the standard library inside the Swift. module.

In general, I would advise to implement your custom logging system and enforce its usage by other means, e.g. code reviews or linting rules (e.g. swiftlint).

But he says in his question “Overriding it is not an option, as it may get bypasses if you use Swift.print(:)”. So he knows this and rejects it. — matt, Jan 01 '19 at 19:21
@matt Against by better judgement I have updated my answer to solve that. — Sulthan, Jan 01 '19 at 19:29
Thanks, @Sulthan , your solution to the problem is interesting. — Tal Zion, Jan 02 '19 at 06:23

score 1 · Answer 2 · edited Jan 01 '19 at 20:59

Adding to the comments by Alexander and Carpsen90

Method Swizzling is an Objective-C runtime feature and it is inherently not available in Swift as Swift is not a dynamic language. However, you can have a global function as in this SO Post. Below is the updated code for Swift 4.2. But unfortunately, it calls the original Swift.print function if the function name is print. So I changed the function name to logs

public func logs(items: Any..., separator: String = " ", terminator: String = "\n") {
    let output = items.map { "*\($0)"}.joined(separator: " ")
    Swift.print(output, terminator: terminator)
}

Another Possible option is to put it into a protocol

public protocol CustomPrintable {
    func print(_ items: Any...)
}

extension CustomPrintable {
    func print(_ items: Any...) {
        let output = items.map { "*\($0)"}.joined(separator: " ")
        Swift.print("****" + output)
    }
}

Usage

class SampleClass : CustomPrintable {

  func printValue() {
      print ("Hello World")
  }
}

But this way the function ceases to be a global function and while using print you have to select the correct method

Output

*****Hello World

You didn't have correct external parameter names. It would have to be `print(_ items: ...)` not `print(items: ...)`. That's why your method did not get called. — Sulthan, Jan 02 '19 at 08:08
@Sulthan Thanks for pointing it out, I made the change but still the function call goes to the Swift.print function — Md. Ibrahim Hassan, Jan 02 '19 at 16:17

How to swizzle Swift.print(items:separator:terminator)

2 Answers2