如何在我的触摸开始事件中仅访问我的触摸之一

Question

我有：

UITouch *touch = [touches anyObject];

    if ([touches count] == 2) {
        //preforming actions                                                             
    }

我想要做的是在 if 语句内询问它，其中两次触摸是分开的。

Answer 1

您可以迭代触摸：

if([touches count] == 2) {
    for(UITouch *aTouch in touches) {
        // Do something with each individual touch (e.g. find its location)
    }
}

编辑：如果您想要找到两个触摸之间的距离，并且您知道恰好有两个触摸，您可以分别抓取每个触摸，然后进行一些数学运算。例子：

float distance;
if([touches count] == 2) {
    // Order touches so they're accessible separately
    NSMutableArray *touchesArray = [[[NSMutableArray alloc] 
                                     initWithCapacity:2] autorelease];
    for(UITouch *aTouch in touches) {
        [touchesArray addObject:aTouch];
    }
    UITouch *firstTouch = [touchesArray objectAtIndex:0];
    UITouch *secondTouch = [touchesArray objectAtIndex:1];

    // Do math
    CGPoint firstPoint = [firstTouch locationInView:[firstTouch view]];
    CGPoint secondPoint = [secondTouch locationInView:[secondTouch view]];
    distance = sqrtf((firstPoint.x - secondPoint.x) * 
                     (firstPoint.x - secondPoint.x) + 
                     (firstPoint.y - secondPoint.y) * 
                     (firstPoint.y - secondPoint.y));
}

Answer 2

Touches 已经是一个数组了。无需将它们复制到另一个数组中 - 只需使用 [touches objectAtIndex:n] 即可访问 touch n。