为什么ArrayDeque类在pollFirst方法中使用按位运算？(Why the ArrayDeque class use bitwise operation in the pollFirst method?)

 我通过java源代码看看尝试学习集合的实现。 在ArrayDeque类中发现了一件有趣的事情。 
 
public E pollFirst() {
    int h = head;
    @SuppressWarnings("unchecked")
    E result = (E) elements[h];
    // Element is null if deque empty
    if (result == null)
        return null;
    elements[h] = null;     // Must null out slot
    head = (h + 1) & (elements.length - 1);
    return result;
}

public E pollLast() {
    int t = (tail - 1) & (elements.length - 1);
    @SuppressWarnings("unchecked")
    E result = (E) elements[t];
    if (result == null)
        return null;
    elements[t] = null;
    tail = t;
    return result;
}
 
 以下2行是什么意思？ 这是一个按位操作吗？ 为什么他们使用它，这里的目的是什么？ 
 
    head = (h + 1) & (elements.length - 1);
    int t = (tail - 1) & (elements.length - 1);
 
 我知道一个使用按位的方案是在1个变量中打包2个值。 但似乎事实并非如此。 

I look through java source code try to learn the implementation of collection. Found a interesting thing in the ArrayDeque class.
 
public E pollFirst() {
    int h = head;
    @SuppressWarnings("unchecked")
    E result = (E) elements[h];
    // Element is null if deque empty
    if (result == null)
        return null;
    elements[h] = null;     // Must null out slot
    head = (h + 1) & (elements.length - 1);
    return result;
}

public E pollLast() {
    int t = (tail - 1) & (elements.length - 1);
    @SuppressWarnings("unchecked")
    E result = (E) elements[t];
    if (result == null)
        return null;
    elements[t] = null;
    tail = t;
    return result;
}
 
What does the following 2 lines mean? Is it a bitwise operation? Why they use it and what's the purpose here?
 
    head = (h + 1) & (elements.length - 1);
    int t = (tail - 1) & (elements.length - 1);
 
I know one scenario to use the bitwise is to pack 2 values in 1 variable. But it seems this is not the case. 

原文：https://stackoverflow.com/questions/36825213