输出：没有错误您可能分配了太多内存 任何人都可以详细解释

Question

查看以下源代码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：没有错误您可能分配了太多内存谁能详细解释一下

Answer 1

将您的代码更改为

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

如果有人仍在寻找为什么 15 位小数精度出现的答案。

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