查找 NSArray 中矩形的最大宽度

发布于 2024-10-31 22:00:22 字数 484 浏览 0 评论 0原文

我正在使用包含 CGRect 值的 NSArray 。有没有办法通过库方法获得数组中包含的所有矩形的最大宽度，或者我是否必须实现自己的逻辑？提前致谢。

代码示例：

for (int i=0; i<[array1 count]; i++) {
    CGRect rect = [[array1 objectAtIndex:i] CGRectValue];
// Here, some transformation of rects

[array2 addObject:[NSValue valueWithCGRect:rect]];

}

在此代码中，我尝试从 array1 获取帧并将其添加到 array2 中。我从 XML 中获取 array1 中的帧。我解析了这些帧并将它们放入数组中。我以一种非常简单的方式提供了我的代码。

原文

I am using an NSArray which contains CGRect values. Is there any way to get the greatest width of all the rects contained in the array through library methods, or do I have to implement my own logic? Thanks in advance.

Code sample:

for (int i=0; i<[array1 count]; i++) {
    CGRect rect = [[array1 objectAtIndex:i] CGRectValue];
// Here, some transformation of rects

[array2 addObject:[NSValue valueWithCGRect:rect]];

}

In this code I am trying to take frames from array1 and add them to array2. I am getting the frames in array1 from XML. I parsed those frames and placed them in the array. I provided my code as a very simple way.

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陪你搞怪i 2024-11-07 22:00:22

不要像这样一遍又一遍地向您的数组发送 -count 消息。这是浪费。

float result = 0.0;
for (value in array1)
  {
  GCRect rect = [value CGRectValue];
  result = MAX(result, rect.size.width);
  }  // "result" now contains the greatest width you saw in the loop.

Don't send a -count message to your array over and over like that. It's wasteful.

float result = 0.0;
for (value in array1)
  {
  GCRect rect = [value CGRectValue];
  result = MAX(result, rect.size.width);
  }  // "result" now contains the greatest width you saw in the loop.

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隐诗 2024-11-07 22:00:22

for(int i=0; i < [array2 count]; i++) 
{
    CGRect rect = [[array2 objectAtIndex:i] CGRectValue];
    float widthTest = rect.size.width;
    //this gives you the width of each element. Compare and get the biggest
}

for(int i=0; i < [array2 count]; i++) 
{
    CGRect rect = [[array2 objectAtIndex:i] CGRectValue];
    float widthTest = rect.size.width;
    //this gives you the width of each element. Compare and get the biggest
}

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