java为什么参数。getType（）。isInstance（HttpServletRequest.class）返回值为false，但use“==”为true

Question

Parameter[] ps = method.getParameters();

Map<String,Integer> map = new HashMap<String,Integer>();

for(int ij = 0;ij<ps.length;ij++){

    Parameter p = ps[ij];

    RequestParam rp = p.getAnnotation(RequestParam.class);

    if(rp != null){

        //do something

    }else {
        System.out.println(p.getType());
        System.out.println(p.getType().isInstance(HttpServletRequest.class));
        System.out.println(p.getType() == HttpServletRequest.class);
    }
}

输出为：

interface javax.servlet.http.HttpServletRequest
false
true

为什么使用“isInstance”是假的而使用“==”是真的？ 因为“实例”不能判断关系

Answer 1

isInstance等于instanceOf

This method is the dynamic equivalent of the Java language instanceof operator.

该方法返回false，因为您正在将一个类（由p.getType（）返回）与另一个类HttpServletRequest进行比较。类。此方法需要一个实例，例如：

Dog bobby = new BobbyDog(); // class BobbyDog extends Dog
System.out.println(Dog.class.isInstance(bobby)); // correct use (return true)
System.out.println(Dog.class.isInstance(BobbyDog.class)); // incorrect use (return false)

equals运算符返回true，因为这两个类相等

p.getType() == HttpServletRequest.class // true
HttpServletRequest.class == HttpServletRequest.class // true

如果你想了解这种关系，你必须使用这种方法

isAssignableFrom(Class<?> cls) 

Determines if the class or interface represented by this Class object is either the same as, or is a superclass or superinterface of, the class or interface represented by the specified Class parameter. It returns true if so; otherwise it returns false. If this Class object represents a primitive type, this method returns true if the specified Class parameter is exactly this Class object; otherwise it returns false.

Answer 2

该类型不是HttpServletRequest类的实例，而是包含HttpServletRequest类信息的^{}实例