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先说动机 , 引用自 https://docs.swift.org/swift-book/LanguageGuide/Functions.html In-Out Parameters 章节:
Function parameters are constants by default. Trying to change the value of a function parameter from within the body of that function results in a compile-time error. This means that you can’t change the value of a parameter by mistake. If you want a function to modify a parameter’s value, and you want those changes to persist after the function call has ended, define that parameter as an in-out parameter instead.

再说行为, 引用自 https://docs.swift.org/swift-book/ReferenceManual/Declarations.html#ID545:
In-out parameters are passed as follows:
1. When the function is called, the value of the argument is copied.
2. In the body of the function, the copy is modified.
3.When the function returns, the copy’s value is assigned to the original argument.
This behavior is known as copy-in copy-out or call by value result. For example, when a computed property or a property with observers is passed as an in-out parameter, its getter is called as part of the function call and its setter is called as part of the function return.

多说几句，编译器有可能会把 inout 优化为所谓的传址(call-by-reference)，但是官方特意指出开发者们不能依赖这个行为:
As an optimization, when the argument is a value stored at a physical address in memory, the same memory location is used both inside and outside the function body. The optimized behavior is known as call by reference;
Write your code using the model given by copy-in copy-out, without depending on the call-by-reference optimization, so that it behaves correctly with or without the optimization

不严谨的验证方式是，传一个带有 willSet 或者 didSet 的变量到一个 function 的 inout 参数内，就算在 function 内不改变值，willSet 或者 didSet 也会触发

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@studytogether 老哥能留个联系方式吗 了解一下具体流程