为什么双参数化函数接受浮点数而不是相反？

Question

I have encountered something strange today. The below code compiles unexpectedly and runs fine.

public class Test {

    public static void func(double d) {
        System.out.print("d : " + d);
    }

    public static void main(String[] args) {
        float f = 10.0f;
        func(f); // output: d : 10.0
    }
}

But this one gives compilation error

public class Test {

    public static void func(float f) {
        System.out.print("f : " + f);
    }

    public static void main(String[] args) {
        double d = 10.0d;
        func(d);
    }
}

Can somebody please explain this behaviour ?

Answer 1

Type promotion from float to double is safe as no data is lost and all float 4 bytes can fit into double 8 byes.

However the opposite, from double to float , always truncates the data as double 8 bytes can't fit into float 4 bytes. Compiler guards against doing this truncation accidentally by forcing the programmer to manually specify the type conversion.

Answer 2

double (8 byte) is a bigger data type than float (4 byte) so you can store float (4 byte) in double (8 byte) but you can't double in float . If you try to do that you'll get Possible loss of precision error.

So the following will give error.

float f = 120.55;

While this one don't

double d = 120.44f;