Question

我正在尝试绘制一个矩形，让用户根据他点击的位置和拖动鼠标的位置创建它。

我用来绘制NSRect的代码是：

CGFloat width = endPoint.x-startPoint.x; //Width of rectangle.
CGFloat height = endPoint.y-startPoint.y; //Height of rectangle.
CGFloat rectXPoint = startPoint.x; //x component of corner of rect
CGFloat rectYPoint = startPoint.x; //y component of corner of rect
if (width < 0) { //Handle negavive rect size here.
    width = startPoint.x-endPoint.x;
    rectXPoint = endPoint.x;
}
if (height < 0) { //and here.
    height = startPoint.y-endPoint.y;
    rectYPoint = endPoint.y;
}
NSRect theRect = NSMakeRect(rectXPoint, rectYPoint, width, height);
[[NSColor blackColor] set];
NSRectFill(theRect);

所以我可以让代码工作但不完全正确。有时它会将光标的y偏移一些看似随意的量（即使我知道它不是任意的）。考虑到这将涉及我将鼠标移动到需要证明这一点的点上，很难得到屏幕截图。万一你需要查看我在这里得到鼠标位置的代码，它是：

- (void)mouseDown:(NSEvent *)event {
    NSPoint location = [self convertPointFromBase:[event locationInWindow]];
    startPoint = location;
}

- (void)mouseDragged:(NSEvent *)event {
    NSPoint location = [self convertPointFromBase:[event locationInWindow]];
    endPoint = location;
    [self setNeedsDisplay:YES];
}

isFlipped返回YES。

如果您需要进一步澄清，请询问您需要澄清的内容。

Answer 1

可能这一行：

CGFloat rectYPoint = startPoint.x; //y component of corner of rect

我无法根据用户点击的位置绘制NSRect

1 个答案: