ios - OptionSetType 协议(protocol) Swift

Question

Swift 的 OptionSetType 协议(protocol)的目的是什么，它与仅仅利用 Set 来获取那些 SetAlgebraType 方法有何不同？

Answer 1

我将从使用OptionSetType时的实际角度来回答。对我来说，OptionSetType 的目的是移除 C/ObjC 中的所有位操作。

考虑一个例子，你有一个绘制矩形边框的函数。用户可以要求它在 0 到 4 个边框之间绘制。

这是 Swift 中的代码:

struct BorderType: OptionSetType {
    let rawValue: Int
    init (rawValue: Int) { self.rawValue = rawValue }

    static let Top = BorderType(rawValue: 1 << 0)
    static let Right = BorderType(rawValue: 1 << 1)
    static let Bottom = BorderType(rawValue: 1 << 2)
    static let Left = BorderType(rawValue: 1 << 3)
}

func drawBorder(border: BorderType) {
    // Did the user ask for a Left border?
    if border.contains(.Left) { ... }

    // Did the user ask for both Top and Bottom borders?
    if border.contains([.Top, .Bottom]) { ... }

    // Add a Right border even if the user didn't ask for it
    var border1 = border
    border1.insert(.Right)

    // Remove the Bottom border, always
    var border2 = border
    border2.remove(.Bottom)
}

drawBorder([.Top, .Bottom])

在 C 中:

typedef enum {
    BorderTypeTop    = 1 << 0,
    BorderTypeRight  = 1 << 1,
    BorderTypeBottom = 1 << 2,
    BorderTypeLeft   = 1 << 3 
} BorderType;

void drawBorder(BorderType border) {
    // Did the user ask for a Left border?
    if (border & BorderTypeLeft) { ... }

    // Did the user ask for both Top and Bottom borders?
    if ((border & BorderTypeTop) && (border & BorderTypeBottom)) { ... }

    // Add a Right border even if the user didn't ask for it
    border |= BorderTypeRight;

    // Remove the Bottom border, always
    border &= ~BorderTypeBottom;
}

int main (int argc, char const *argv[])
{
    drawBorder(BorderTypeTop | BorderTypeBottom);
    return 0;
}

C 比较短，但一眼就能看出这行是干什么的？

border &= ~BorderTypeBottom;

虽然这很有意义:

var border2 = border
border2.remove(.Bottom)

与 C 的斯巴达哲学相比，它符合 Swift 的目标，即成为一种富有表现力、易于学习的语言。