java - 为什么三元运算符会意外转换整数？

Question

我在某处看到它讨论过以下代码导致 obj 成为 Double，但它从左侧打印 200.0一边。

Object obj = true ? new Integer(200) : new Double(0.0);

System.out.println(obj);

结果:200.0

---

但是，如果您将不同的对象放在右侧，例如BigDecimal，obj的类型应该是Integer。

Object obj = true ? new Integer(200) : new BigDecimal(0.0);

System.out.println(obj);

结果:200

---

我认为这是因为将左侧转换为 double 与 integer/ 发生的方式相同双重比较和计算，但是这里左右两边不这样交互。

为什么会这样？

Answer 1

您需要阅读 section 15.25 of the Java Language Specification .

特别是:

Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:

• If one of the operands is of type byte or Byte and the other is of type short or Short, then the type of the conditional expression is short.
• If one of the operands is of type T where T is byte, short, or char, and the other operand is a constant expression of type int whose value is representable in type T, then > - the type of the conditional expression is T.
• If one of the operands is of type Byte and the other operand is a constant expression of type int whose value is representable in type byte, then the type of the conditional expression is byte.
• If one of the operands is of type Short and the other operand is a constant expression of type int whose value is representable in type short, then the type of the conditional expression is short.
• If one of the operands is of type; Character and the other operand is a constant expression of type int whose value is representable in type char, then the type of the conditional expression is char.
• Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands. Note that binary numeric promotion performs unboxing conversion (§5.1.8) and value set conversion (§5.1.13).

所以binary numeric promotion已应用，开头为:

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:

• If any of the operands is of a reference type, unboxing conversion (§5.1.8) is performed. Then:
• If either operand is of type double, the other is converted to double.

这正是这里发生的 - 参数类型分别转换为 int 和 double，然后第二个操作数(原始表达式中的第三个)的类型为 double，所以整体结果类型是double。