Arrays.asList得到的list无法在进行add操作

电玩女神 2023-07-02 05:28 17阅读 0赞

List name = Arrays.asList(“xxx”,“yyy”,“zzz”);

***这样的话这个list的size就固定了，不能再add了，要注意。***

为啥呢？参考：  
https://www.cnblogs.com/one-lightyear/p/7513013.html

Arrays.asList生成的ArrayList是自己写的，继承了AbstractList但是没有实现add的方法……  
所以是不行的，AbstractList的add一被调用就会报错

private static class ArrayList<E> extends AbstractList<E>
        implements RandomAccess, java.io.Serializable
    { 
        private static final long serialVersionUID = -2764017481108945198L;
        private final E[] a;
    
        ArrayList(E[] array) { 
            a = Objects.requireNonNull(array);
        }
    
        @Override
        public int size() { 
            return a.length;
        }
    
        @Override
        public Object[] toArray() { 
            return a.clone();
        }
    
        @Override
        @SuppressWarnings("unchecked")
        public <T> T[] toArray(T[] a) { 
            int size = size();
            if (a.length < size)
                return Arrays.copyOf(this.a, size,
                                     (Class<? extends T[]>) a.getClass());
            System.arraycopy(this.a, 0, a, 0, size);
            if (a.length > size)
                a[size] = null;
            return a;
        }
    
        @Override
        public E get(int index) { 
            return a[index];
        }
    
        @Override
        public E set(int index, E element) { 
            E oldValue = a[index];
            a[index] = element;
            return oldValue;
        }
    
        @Override
        public int indexOf(Object o) { 
            E[] a = this.a;
            if (o == null) { 
                for (int i = 0; i < a.length; i++)
                    if (a[i] == null)
                        return i;
            } else { 
                for (int i = 0; i < a.length; i++)
                    if (o.equals(a[i]))
                        return i;
            }
            return -1;
        }
    
        @Override
        public boolean contains(Object o) { 
            return indexOf(o) != -1;
        }
    
        @Override
        public Spliterator<E> spliterator() { 
            return Spliterators.spliterator(a, Spliterator.ORDERED);
        }
    
        @Override
        public void forEach(Consumer<? super E> action) { 
            Objects.requireNonNull(action);
            for (E e : a) { 
                action.accept(e);
            }
        }
    
        @Override
        public void replaceAll(UnaryOperator<E> operator) { 
            Objects.requireNonNull(operator);
            E[] a = this.a;
            for (int i = 0; i < a.length; i++) { 
                a[i] = operator.apply(a[i]);
            }
        }
    
        @Override
        public void sort(Comparator<? super E> c) { 
            Arrays.sort(a, c);
        }
    }

suList也有类似的问题，内部类  
https://www.jianshu.com/p/69937e239d61

subList 返回仅仅只是一个视图而已

它还是原list的引用。