简化 NSBezierPath（或 UIBezierPath）代码，如何？答案

【问题标题】：Simplifying NSBezierPath (Or UIBezierPath) code, How?简化 NSBezierPath（或 UIBezierPath）代码，如何？
【发布时间】：2016-02-21 10:00:13
【问题描述】：

theLen = [NSBezierPath bezierPath];
[theLen moveToPoint:NSMakePoint(_frame.size.width/2 +  base / 2 , _frame.size.height/2 - (diameter / 2))];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 -  base / 2 , _frame.size.height/2 - (diameter / 2))];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 - (thickness / 2 ) , _frame.size.height/2)];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 -  base / 2 , _frame.size.height/2 + (diameter / 2))];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 + base / 2 , _frame.size.height/2 + (diameter / 2))];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 + (thickness / 2 ) , _frame.size.height/2)];
[theLen closePath];
theLen.lineWidth = 2;
[theLen stroke];
[theLen fill];

鉴于此代码，您将如何简化它？我正在尝试学习如何简化代码，而我所能想到的就是把它变成......

float commonNSPoint[4];
commonNSPoint[0] = _frame.size.width/2 +  base / 2;
commonNSPoint[1] = _frame.size.width/2 -  base / 2;
commonNSPoint[2] = _frame.size.height/2 - (diameter / 2);
commonNSPoint[3] = _frame.size.height/2 + (diameter / 2);

theLen = [NSBezierPath bezierPath];
[theLen moveToPoint:NSMakePoint( commonNSPoint[0],  commonNSPoint[2])];
[theLen lineToPoint:NSMakePoint(commonNSPoint[1],  commonNSPoint[2])];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 - (thickness / 2 ) , _frame.size.height/2)];
[theLen lineToPoint:NSMakePoint(commonNSPoint[1],  commonNSPoint[3])];
[theLen lineToPoint:NSMakePoint( commonNSPoint[0], commonNSPoint[3])];
[theLen lineToPoint:NSMakePoint(_frame.size.width/2 + (thickness / 2 ) , _frame.size.height/2)];
[theLen closePath];
[theLen closePath];
theLen.lineWidth = 2;
[theLen stroke];
[theLen fill];

这并没有真正帮助我，因为 NSBezierPath 或 UIBezierPath 的点位置并不完全相同。这也使代码看起来更混乱。

你们将如何简化这样的 NSBezierPath 或 UIBezierPath。任何事情都会有所帮助（因为我的程序基于不同的 NSPoint 值和许多 NSBezierPath 绘图）。

谢谢

【问题讨论】：

标签： objective-c uibezierpath nsbezierpath

【解决方案1】：

首先，分解出你的常用子表达式并给它们起有意义的名字（commonNSPoint 不是）。这是一个合理的开始：

    CGFloat xMid = _frame.size.width / 2;
    CGFloat yMid = _frame.size.height / 2;
    CGFloat radius = diameter / 2;
    CGFloat halfBase = base / 2;
    CGFloat halfThick = thickness / 2;

然后制作一个仅包含点的数组，而不需要函数和方法调用将事情弄得一团糟：

    CGPoint points[] = {
        { xMid + halfBase,  yMid - radius },
        { xMid - halfBase,  yMid - radius },
        { xMid - halfThick, yMid },
        { xMid - halfBase,  yMid + radius },
        { xMid + halfBase,  yMid + radius },
        { xMid + halfThick, yMid },
    };

最后，创建一个包含这些点的路径：

    theLen = [NSBezierPath bezierPath];
    [theLen appendBezierPathWithPoints:points count:sizeof points / sizeof points[0]];
    [theLen closePath];

或者，让点以原点为中心，然后在创建路径后平移（滑动）路径：

    CGFloat radius = diameter / 2;
    CGFloat halfBase = base / 2;
    CGFloat halfThick = thickness / 2;

    CGPoint points[] = {
        { halfBase,  radius },
        { halfBase,  radius },
        { halfThick, 0 },
        { halfBase,  radius },
        { halfBase,  radius },
        { halfThick, 0 },
    };

    theLen = [NSBezierPath bezierPath];
    [theLen appendBezierPathWithPoints:points count:sizeof points / sizeof points[0]];
    [theLen closePath];

    NSAffineTransform *transform = [NSAffineTransform transform];
    [transform translateXBy:_frame.size.width / 2 yBy:_frame.size.height / 2];
    [theLen transformUsingAffineTransform:transform];

【讨论】：