输出：无错误您可能分配了过多的内存，任何人都可以详细解释

Question

查看下面的源代码

class Test
{
    public static void main(String[] args)
    {
        Double object = new Double("2.4");
        int a = object.intValue();
        byte b = object.byteValue();
        float d = object.floatValue();
        double c = object.doubleValue();

        System.out.println(a + b + c + d );

    }
}

  

输出：无   错误   您可能分配了过多的内存

谁能详细解释

Answer 1

将代码更改为

Double object = new Double("2.4");
int a = object.intValue();
byte b = object.byteValue();
float d = object.floatValue();
double c = object.doubleValue();

System.out.println(a);        
System.out.println(a + b);        
System.out.println(a + b + c);        
System.out.println(a + b + c + d );   

并探索

Answer 2

万一有人还在寻找答案，为什么会有15位小数的精度。

public class JavaExample {

    public static void main(String[] args) {
        Double object = new Double("2.4");
        int a = object.intValue();
        byte b = object.byteValue();
        float c = object.floatValue();
        double d = object.doubleValue();
        float e = object.floatValue();
        double f = object.doubleValue();

        System.out.println("int :"+a);        
        System.out.println("int and byte addition :"+(a + b));        
        System.out.println("int, byte and float addition :"+(a + b + d));        
        System.out.println("int, byte, float and double addition :"+(a + b + c + d));   
        System.out.println("float and double addition :"+(c + d));  
        System.out.println("float and float addition :"+(c + e)); 
        System.out.println("double and double addition :"+(d + f)); 
        System.out.println("float and double addition :"+(e + f)); 
    }
}
Output:
int :2
int and byte addition :4
int, byte and float addition :6.4
int, byte, float and double addition :8.800000095367432
float and double addition :4.800000095367432
float and float addition :4.8
double and double addition :4.8
float and double addition :4.800000095367432