将表达式的结果分配给基元 [英] Assigning result of an expression to a primitive

问题描述

K.Sierra在她的书SCJP学习指南中提到我们知道一个文字整数总是一个int，但更重要的是，涉及任何int-sized或更小的表达式的结果总是一个int 。

K.Sierra in her book "SCJP Study Guide" mentions "We know that a literal integer is always an int, but more importantly, the result of an expression involving anything int-sized or smaller is always an int."

我已经开始尝试了，我对以下结果感到有些困惑：

I've started experimenting and I'm a little bit confused with the below results:

byte a = 1; // correct
byte b = 1 + a; // incorrect (needs explicit casting)
byte c = 1 + 1; // correct (I expected it to be incorrect)

有人可以向我解释为什么最后一个例子做了不需要铸造？为什么Java编译器会进行隐式转换？是因为有2个文字？澄清非常感谢。

Could anyone explain to me why the last example does not require casting? Why does the Java compiler put the implicit cast? Is it because there are 2 int literals? Clarification greatly appreciated.

推荐答案

隐式类型转换只有在知道 RHS 的值时才有效在编译时，表示它们是编译时常量。在其他情况下，您需要进行明确的类型转换。

Implicit typecasting only works when the value of your RHS is known at compile-time, means they are compile-time constants. In other cases, you need to do explicit typecasting.

所以： -

byte c = 1 + 1; // Value of `1 + 1` is known at compile time. Implicit cast
byte c = 1 + a; // Value of `1 + a` is evaluated at runtime. Explicit cast needed

另外，请注意，如果你声明你的变量 a as 最后一个字节a = 1 ，然后第二个作业将编译，就像那个案例一样，你的 a 将是编译时常量。

Also, note that, if you declare your variable a as final byte a = 1, then the 2nd assignment will compile, as in that case, your a will be a compile time constant.

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