String "+"的问题

Java String拼接与null值

最新推荐文章于 2024-09-07 17:58:41 发布

原创最新推荐文章于 2024-09-07 17:58:41 发布 · 422 阅读

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本文探讨了Java中String拼接操作时遇到的空指针问题，并详细解释了String类+运算背后的机制，包括如何处理null字符串。

今天项目查bug时,发现一个空指针问题,查了半天,最终定位到是有String +运算造成的.于是就写了一个测试:

public static void main(String [] arg0){
        String a = null;
        String b = "null";
        String c = "===";
        String  ac = a+c;
        String  bc = b+c;
        System.out.println("a+c: " + ac);
        System.out.println("b+c: " + bc);
        System.out.println("a+c == b+c " + ac.equals(bc));
    }

运行结果:

a+c: null===
b+c: null===
a+c == b+c true

原因: Java只是不支持自定义运算符重载，这个String类的+就是运算符重载，也是Java中仅有的一个运算符重载，它是通过编译器实现的，在语法解析阶段就已经确定了.

如:a+c是通过 new StringBuilder().append(a).append(c)实现的.

@Override
public StringBuilder append(String str) {
    super.append(str);
    return this;
}

/**
     * Appends the specified string to this character sequence.
     * <p>
     * The characters of the {@code String} argument are appended, in
     * order, increasing the length of this sequence by the length of the
     * argument. If {@code str} is {@code null}, then the four
     * characters {@code "null"} are appended.
     * <p>
     * Let <i>n</i> be the length of this character sequence just prior to
     * execution of the {@code append} method. Then the character at
     * index <i>k</i> in the new character sequence is equal to the character
     * at index <i>k</i> in the old character sequence, if <i>k</i> is less
     * than <i>n</i>; otherwise, it is equal to the character at index
     * <i>k-n</i> in the argument {@code str}.
     *
     * @param   str   a string.
     * @return  a reference to this object.
     */
public AbstractStringBuilder append(String str) {
    if (str == null)
        return appendNull();
    int len = str.length();
    ensureCapacityInternal(count + len);
    str.getChars(0, len, value, count);
    count += len;
    return this;
}