java - getMethod 的第二个参数是null和new Class[]{}的区别？

Question

在反射方法的时候，如果某方法没有参数好像有下面两种方法去获得 getMethod(method_name_str, new Class[]{});或者getMethod(method_name_str, null); 请问这两种方法有区别吗？

高洛峰 · Answer

In terms of effect, after Class 类的 getMethod 方法，这两种参数没有区别。
我们可以查看 getMethod 的源码，getMethod is called layer by layer, the following method will be used:

In this method, you can see that there is a arrayContentsEq method used to match the parameters of the method:

It can be found that for the case where parameterTypes is null, and for the case where null 的情况，和对于 parameterTypes 为空数组（length == 0）的情况，效果是一样的 —— 假设此时我们要获取的方法 m 的参数为空，那么该方法的 m.getParameterTypes() 返回的数组（a2）的长度即为 0，我们可以发现 a1 == null 或者 a1.length == 0 的时候，arrayContentsEq 方法返回的都是 trueparameterTypes

is an empty array (length == 0), the effect is the same - assuming that we want to obtain at this time If the parameters of method m are empty, then the length of the array (a2) returned by m.getParameterTypes() of this method is 0. We can find that a1 == null or When a1.length == 0, the 🎜 method returns true (that is, the match is successful). 🎜

大家讲道理 · Answer

If a method has no parameters, there is actually no difference between the two situations.

TrackgetMethod(String name, Class... parameterTypes)的源码，可以发现如下代码，其中a1为传入的parameterTypes，a2为根据参数name找到的Method实例调用的method.getParameterTypes()。程序根据比较a1和a2来返回正确的Method.

private static boolean arrayContentsEq(Object[] a1, Object[] a2) {
    if (a1 == null) {
        return a2 == null || a2.length == 0;
    }

    if (a2 == null) {
        return a1.length == 0;
    }

    if (a1.length != a2.length) {
        return false;
    }

    for (int i = 0; i < a1.length; i++) {
        if (a1[i] != a2[i]) {
            return false;
        }
    }

    return true;
}