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

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?

天蓬老师天蓬老师2804 days ago999

reply all(2)I'll reply

  • 高洛峰

    高洛峰2017-04-18 10:50:53

    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). 🎜

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  • 大家讲道理

    大家讲道理2017-04-18 10:50:53

    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parameterTypesa2为根据参数name找到的Method实例调用的method.getParameterTypes()。程序根据比较a1a2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;
    }

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