查看以下源代碼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 ); }}輸出：沒有錯誤您可能分配了太多內(nèi)存誰能詳細(xì)解釋一下
2 回答
慕的地10843
TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個贊
將您的代碼更改為
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
繁星淼淼
TA貢獻(xiàn)1775條經(jīng)驗(yàn) 獲得超11個贊
如果有人仍在尋找為什么 15 位小數(shù)精度出現(xiàn)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
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