具有最终访问修饰符的char和byte-Java [英] char and byte with final access modifier - java

问题描述

请看下面的示例，我无法理解字符和字节之间的关系

Please take a look at below example i cant understand the relation between char and byte

byte b = 1;
char c = 2;

c = b; // line 1

给我编译错误，因为c为char类型，b为byte类型，因此强制转换必须处于这种状态

Give me compilation Error because c is type of char and b is type of byte so casting is must in such condition

但是现在这里的tweest是当我在代码下运行时

but now the tweest here is when i run below code

final byte b = 1;
char c = 2;

c = b; // line 2

第2行编译成功，根本不需要任何强制转换 所以我的问题是为什么当我在byte

line 2 compile successfully it doesn't need any casting at all so my question is why char c behave different when i use final access modifier with byte

推荐答案

由于使用final对其进行了限定，因此使该变量成为常量变量，这是

Because qualifying it with final makes the variable a constant variable, which is a constant expression. So

final byte b = 1;
char c = 2;

c = b; // line 2

实际上变成了

final byte b = 1;
char c = 2;

c = 1;

并且编译器保证值1可以适合char变量.

And the compiler has a guarantee that the value 1 can fit in a char variable.

对于非常量byte变量，没有这样的保证. 已签名byte，未签名char .

With a non constant byte variable, there is no such guarantee. byte is signed, char is unsigned.

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