Java Puzzler - 将double转换为int [英] Java Puzzler - casting a double to int

问题描述

int anInt = 1;
double aDouble = 2.5;

anInt = anInt + aDouble; // Error - need to cast double to int

anInt += aDouble; // This is ok. Why?

anInt = aDouble; // This is also an error.

anInt = 1 + aDouble; // This is also an error.

所以我的问题是：为什么 anInt不是编译错误+ = aDouble ？

So my questions is: Why is it not a compile error to do anInt += aDouble?

推荐答案

四个案例中有三个正确报告错误。复合赋值是规则的唯一例外。 Java语言规范，第15.26.2节，解释了原因：

Three of the four cases properly report an error. Compound assignment is the only exception from the rule. Java Language Specification, part 15.26.2, explains why:

15.26.2复合赋值运算符

表格 E1 op = E2 的复合赋值表达式相当于 E1
=（T）（（E1）op（E2）），其中 T 是 E1 ，但 E1 仅评估一次。

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

例如，以下代码是正确的：

For example, the following code is correct:

short x = 3;
x += 4.6;

并导致x的值为7，因为它相当于：

and results in x having the value 7 because it is equivalent to:

short x = 3;
x = (short)(x + 4.6);

如您所见，隐式插入可以避免错误演员。

As you can see, the error is avoided by implicit insertion of a cast.

这篇关于Java Puzzler - 将double转换为int的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！