复合分配的自动（取消）装箱失败

发布于 2024-08-30 06:50:39 字数 828 浏览 4 评论 0原文

由于复合赋值和递增/递减运算符中的隐式转换，以下代码可以编译：

byte b = 0;
++b; b++; --b; b--;
b += b -= b *= b /= b %= b;
b <<= b >>= b >>>= b;
b |= b &= b ^= b;

并且由于自动装箱和自动拆箱，以下代码也可以编译：

Integer ii = 0;
++ii; ii++; --ii; ii--;
ii += ii -= ii *= ii /= ii %= ii;
ii <<= ii >>= ii >>>= ii;
ii |= ii &= ii ^= ii;

然而，以下代码片段中的最后一行给出了编译时错误：

Byte bb = 0;
++bb; bb++; --bb; bb--; // ... okay so far!
bb += bb; // DOESN'T COMPILE!!!
// "The operator += is undefined for the argument type(s) Byte, byte"

谁能帮我弄清楚这里发生了什么事吗？ byte b 版本编译得很好，所以 Byte bb 不应该效仿并根据需要进行适当的装箱和拆箱以适应吗？

额外问题

那么有没有办法让复合赋值运算符与左侧的 Byte、Character 和 Short 一起使用，或者它们是对于这些类型来说根本不合法（！！！）？

原文

Thanks to the implicit casting in compound assignments and increment/decrement operators, the following compiles:

byte b = 0;
++b; b++; --b; b--;
b += b -= b *= b /= b %= b;
b <<= b >>= b >>>= b;
b |= b &= b ^= b;

And thanks to auto-boxing and auto-unboxing, the following also compiles:

Integer ii = 0;
++ii; ii++; --ii; ii--;
ii += ii -= ii *= ii /= ii %= ii;
ii <<= ii >>= ii >>>= ii;
ii |= ii &= ii ^= ii;

And yet, the last line in the following snippet gives compile-time error:

Byte bb = 0;
++bb; bb++; --bb; bb--; // ... okay so far!
bb += bb; // DOESN'T COMPILE!!!
// "The operator += is undefined for the argument type(s) Byte, byte"

Can anyone help me figure out what's going on here? The byte b version compiles just fine, so shouldn't Byte bb just follow suit and do the appropriate boxing and unboxing as necessary to accommodate?

Extra question

So is there a way to make compound assignment operators work with Byte, Character, and Short on the left hand side, or are they simply illegal(!!!) for these types?

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惟欲睡 2024-09-06 06:50:39

§ 5.1.7（拳击）节标准的内容是：

From type boolean to type Boolean
From type byte to type Byte
From type char to type Character
From type short to type Short
From type int to type Integer
From type long to type Long
From type float to type Float
From type double to type Double

注意没有int to Byte。当您执行 bb + bb 时，它会转换为 int + int，而不会装回 Byte。对于 byte 版本，int + int 直接转换回 byte（缩小原始转换，§ 5.1.3），因此这是允许的。

Section § 5.1.7 (Boxing) of the standard says:

From type boolean to type Boolean
From type byte to type Byte
From type char to type Character
From type short to type Short
From type int to type Integer
From type long to type Long
From type float to type Float
From type double to type Double

Note there is no int to Byte. When you do bb + bb it's converted to int + int, which isn't boxed back to Byte. For the byte version, int + int is cast back to byte directly (narrowing primitive conversions, § 5.1.3) so it's allowed.

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