如何从控件数组中获得面积最小的控件?

你可以用可枚举。聚合为了这个。唉,这个函数会一遍又一遍地计算最小控件的大小。

Controls 并返回最小的一个,与所有其他linq函数类似。看到了吗 extension methods demystified

// extension function: gets the size of a control
public static int GetSize(this Control control)
{
    return (control.Height * control.Width);
}

// returns the smallest control, by size, or null if there isn't any control
public static Control SmallestControlOrDefault(this IEnumerable<Control> controls)
{
    if (controls == null || !controls.Any()) return null; // default

    Control smallestControl = controls.First();
    int smallestSize = smallestControl.GetSize();

    // check all other controls if they are smaller
    foreach (var control in controls.Skip(1))
    {
        int size = control.GetSize();
        if (size < smallestSize)
        {
             smallestControl = control;
             smallestSize = size;
        }
    }
    return smallestControl;
}

Skip(1) 将再次迭代第一个元素。如果您不想这样做,请使用 GetEnumerator MoveNext :

var enumerator = controls.GetEnumerator();
if (enumerator.MoveNext())
{   // there is at least one control
    Control smallestControl = enumerator.Current;
    int smallestSize = smallestControl.GetSize();

    // continue enumerating the other controls, to see if they are smaller
    while (enumerator.MoveNext())
    {   // there are more controls
        Control control = enumerator.Current;
        int size = control.GetSize();
        if (size < smallestSize)
        {   // this control is smaller
            smallestControl = control;
            smallestSize = size;
        }
    }
    return smallestControl;
}
else
{   // empty sequence
    ...
}

用法:

IEnumerable<Control> controls = ...
Control smallestControl = controls.SmallestControlOrDefault();

这要归功于非常强大但未充分利用的 Aggregate

var min = controls.Aggregate((x, y) => x.Height * x.Width < y.Height * y.Width ? x : y);

你可以延长你的时间 Control 用一个 Area 属性以避免 方法。