jdk源码解析(一)---java.long.object

jdk源码解析(一)—java.long.object

我们都知道java中的类都继承自Object类，我们今天就要来探讨下Object类的一些特性，首先我们来看下Object里面有哪些方法

• hashCode()
• equals()
• toString()
• notify()
• notifyAll()
• wait(long, int)
• wait()

我们接下逐渐介绍上面的几个方法

hashCode()

1.String.hashCode

    public int hashCode() {
        int h = hash;
        if (h == 0 && value.length > 0) {
            char val[] = value;

            for (int i = 0; i < value.length; i++) {
                h = 31 * h + val[i];
            }
            hash = h;
        }
        return h;
    }

举例子:
“abc”.hashCode() = 96354;
第一步： a = 97;
第二步: ab = 97*31 + 98 = 3105
第三步: abc = 3105 * 31 + 99 = 96354
2.Integer.hashCode

    public int hashCode() {
        return Integer.hashCode(value);
    }

    public static int hashCode(int value) {
        return value;
    }

3.AbstractList.hashCode

    public int hashCode() {
        int hashCode = 1;
        for (E e : this)
            hashCode = 31*hashCode + (e==null ? 0 : e.hashCode());
        return hashCode;
    }

4.AbstractMap.hashCode

	public int hashCode() {
        int h = 0;
        Iterator> i = entrySet().iterator();
        while (i.hasNext())
            h += i.next().hashCode();
        return h;
    }
    
    public int hashCode() {
        return (key   == null ? 0 :   key.hashCode()) ^
               (value == null ? 0 : value.hashCode());

5.AbstractSet.hashCode

    public int hashCode() {
        int h = 0;
        Iterator i = iterator();
        while (i.hasNext()) {
            E obj = i.next();
            if (obj != null)
                h += obj.hashCode();
        }
        return h;
    }

介绍了这么多hashCode方法，我们来介绍下equal方法，我们都知道，equals和hashCode有着很密切的关系，equals相同，hashCode一定相同，反之，则不是这样,上面我们已经介绍完了equals方法，那接下来我们来介绍下equals方法

equals方法

1.String.equals

    public boolean equals(Object anObject) {
        if (this == anObject) {
            return true;
        }
        if (anObject instanceof String) {
            String anotherString = (String)anObject;
            int n = value.length;
            if (n == anotherString.value.length) {
                char v1[] = value;
                char v2[] = anotherString.value;
                int i = 0;
                while (n-- != 0) {
                    if (v1[i] != v2[i])
                        return false;
                    i++;
                }
                return true;
            }
        }
        return false;
    }

2.Integer.equals

    public boolean equals(Object obj) {
        if (obj instanceof Integer) {
            return value == ((Integer)obj).intValue();
        }
        return false;
    }

3.AbstractList.equals

    public boolean equals(Object o) {
        if (o == this)
            return true;
        if (!(o instanceof List))
            return false;

        ListIterator e1 = listIterator();
        ListIterator e2 = ((List) o).listIterator();
        while (e1.hasNext() && e2.hasNext()) {
            E o1 = e1.next();
            Object o2 = e2.next();
            if (!(o1==null ? o2==null : o1.equals(o2)))
                return false;
        }
        return !(e1.hasNext() || e2.hasNext());
    }

4.AbstractMap.equals

    public boolean equals(Object o) {
        if (o == this)
            return true;

        if (!(o instanceof Map))
            return false;
        Map m = (Map) o;
        if (m.size() != size())
            return false;

        try {
            Iterator> i = entrySet().iterator();
            while (i.hasNext()) {
                Entry e = i.next();
                K key = e.getKey();
                V value = e.getValue();
                if (value == null) {
                    if (!(m.get(key)==null && m.containsKey(key)))
                        return false;
                } else {
                    if (!value.equals(m.get(key)))
                        return false;
                }
            }
        } catch (ClassCastException unused) {
            return false;
        } catch (NullPointerException unused) {
            return false;
        }

        return true;
    }

5.AbstractSet.equals

    public boolean equals(Object o) {
        if (o == this)
            return true;

        if (!(o instanceof Set))
            return false;
        Collection c = (Collection) o;
        if (c.size() != size())
            return false;
        try {
            return containsAll(c);
        } catch (ClassCastException unused)   {
            return false;
        } catch (NullPointerException unused) {
            return false;
        }
    }

toString方法显而易见，就是一个展示的接口，直接看代码吧

toString

1.String.toString

    public String toString() {
        return this;
    }

2.AbstractCollection.toString

    public String toString() {
        Iterator it = iterator();
        if (! it.hasNext())
            return "[]";

        StringBuilder sb = new StringBuilder();
        sb.append('[');
        for (;;) {
            E e = it.next();
            sb.append(e == this ? "(this Collection)" : e);
            if (! it.hasNext())
                return sb.append(']').toString();
            sb.append(',').append(' ');
        }
    }

3.AbstractMap.toString

    public String toString() {
        Iterator> i = entrySet().iterator();
        if (! i.hasNext())
            return "{}";

        StringBuilder sb = new StringBuilder();
        sb.append('{');
        for (;;) {
            Entry e = i.next();
            K key = e.getKey();
            V value = e.getValue();
            sb.append(key   == this ? "(this Map)" : key);
            sb.append('=');
            sb.append(value == this ? "(this Map)" : value);
            if (! i.hasNext())
                return sb.append('}').toString();
            sb.append(',').append(' ');
        }
    }

4.AbstractSet.toString实现方法和AbstractList.toString一样