为什么在调用构造函数时将 int 隐式转换为 float 而不是 double？

Question

我写了这个测试代码

public class ConstructorTestApplication {

    private static String result;


    public static void main(String[] args) {

        ConstructorTest test1 = new ConstructorTest(0);
        System.out.println(result);
    }

    private static class ConstructorTest {

        public ConstructorTest(double param){
            result = "double constructor called!";
        }
        public ConstructorTest(float param) {
            result = "float constructor called!";
        }


    }
}

结果是

float constructor called!

为什么调用了 float 构造函数而不是 double 构造函数？这部分是动态方法查找吗？

Answer 1

ConstructorTest(float param) 是两个构造函数中最具体的方法，因为带有 double 参数的方法可以接受任何 float 值，但反之则不然。

JLS 15.12.2.5:

15.12.2.5. Choosing the Most Specific Method

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

The informal intuition is that one method is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time type error.

Answer 2

我送JLS 5.3. Method Invocation Conversion and 5.1.2. Widening Primitive Conversion

19 specific conversions on primitive types are called the widening primitive conversions:

byte to short, int, long, float, or double

short to int, long, float, or double

char to int, long, float, or double

int to long, float, or double

long to float or double

float to double

本质上，当转换完成时，在方法重载中，浮点数优先于双精度数。

为什么在调用构造函数时将 int 隐式转换为 float 而不是 double？

Why is int implicitly cast to float rather than double when invoking constructor?

java

constructor