toString()方法

时间:2024-02-29 12:33:45

在写toString的时候发现了一个小问题。

      char [] chars={'a','A','c','5','8'};
        int [] ints={1,2,3,4};
        System.out.print(chars.toString()); //无法输出数组中的值
        System.out.print(ints.toString());  

在打印时打印的都是@xxx,不能直接输出里面的值,于是看了眼源码:

    /**
     * Returns a string representation of the object. In general, the
     * {@code toString} method returns a string that
     * "textually represents" this object. The result should
     * be a concise but informative representation that is easy for a
     * person to read.
     * It is recommended that all subclasses override this method.
     * <p>
     * The {@code toString} method for class {@code Object}
     * returns a string consisting of the name of the class of which the
     * object is an instance, the at-sign character `{@code @}', and
     * the unsigned hexadecimal representation of the hash code of the
     * object. In other words, this method returns a string equal to the
     * value of:
     * getClass().getName() + '@' + Integer.toHexString(hashCode())
     * 
     *
     * @return  a string representation of the object.
     */
    public String toString() {
        return getClass().getName() + "@" + Integer.toHexString(hashCode());
    }

这个方法跳到了Object类的toString()方法中,意思是返回的对象的字符串表示形式。Object类的toString方法返回一个字符串,该字符串由对象作为实例的类名(数组也是对象)、@符号字符' @'和对象哈希码的无符号十六进制表示形式组成。

因此我们最后得到的是@和对象的散列值。

而在需要打印值时,可以用:Arrays.toString()方法。

        char [] chars={'a','A','c','5','8'};
        int [] ints={1,2,3,4};
        System.out.print(Arrays.toString(ints));
       System.out.print(Arrays.toString(chars));

Arrays.toString()重写了toString方法,通过StringBuilder使我们拿到了数组中的值,上个源码:

    public static String toString(int[] a) {
        if (a == null)
            return "null";
        int iMax = a.length - 1;
        if (iMax == -1)
            return "[]";

        StringBuilder b = new StringBuilder();
        b.append('[');
        for (int i = 0; ; i++) {
            b.append(a[i]);
            if (i == iMax)
                return b.append(']').toString();
            b.append(", ");
        }

至于为什么使用StringBuilder,因为这显然是个单线程的过程,所以选择能性能更快的StringBuilder.