ios - 在 Swift 中将 nil 分配给泛型可选值

Question

我在 Swift 中创建了一个简单的通用网格数据结构，如下所示。基本上它创建了一个类型为 T 的可选数组？并用 nil 初始化数组。但是，当我尝试将一个网格元素显式设置回 nil 时，编译器会提示一些我不太理解的事情。

struct Grid<T> {
    let columns: Int, rows: Int
    var grid: [T?]
    init(columns: Int, rows: Int) {
        self.rows = rows
        self.columns = columns
        grid = Array(count: rows * columns, repeatedValue: nil)
    }

    func test() {
        grid[0] = nil
    }
}

添加 test() 函数时编译器的抗议:

Grid.swift:26:13: '@lvalue $T7' is not identical to 'T?'

Answer 1

错误信息具有误导性。 test() 方法修改属性的值在结构中，因此您必须将其标记为“变异”:

struct Grid<T> {
    // ...

    mutating func test() {
        grid[0] = nil
    }
}

参见 Modifying Value Types from Within Instance Methods在 Swift 书中:

Modifying Value Types from Within Instance Methods

Structures and enumerations are value types. By default, the properties of a value type cannot be modified from within its instance methods.

However, if you need to modify the properties of your structure or enumeration within a particular method, you can opt in to mutating behavior for that method. ...