为什么我不能添加两个字节并获得一个 int 而我可以添加两个最终字节获得一个字节? [英] Why can not I add two bytes and get an int and I can add two final bytes get a byte?

问题描述

public class Java{
    public static void main(String[] args){
        final byte x = 1;
        final byte y = 2;
        byte z = x + y;//ok
        System.out.println(z);

        byte a = 1;
        byte b = 2;
        byte c = a + b; //Compiler error
        System.out.println(c);
    }
}

如果涉及任何整数或更小的表达式的结果始终是整数，即使两个字节的总和适合一个字节.

If the result of an expression involving anything int-sized or smaller is always an int even if the sum of two bytes fit in a byte.

当我们添加两个适合一个字节的最后一个字节时，为什么会发生这种情况?没有编译器错误.

推荐答案

来自 JLS 5.2 赋值转换

此外，如果表达式是 byte、short、char 或 int 类型的常量表达式(第 15.28 节):- 如果类型为变量是 byte、short 或 char，以及常量的值表达式可以用变量的类型来表示.

In addition, if the expression is a constant expression (§15.28) of type byte, short, char, or int: - A narrowing primitive conversion may be used if the type of the variable is byte, short, or char, and the value of the constant expression is representable in the type of the variable.

简而言之，表达式的值(在编译时已知，因为它是一个常量表达式)可以用字节类型的变量来表示.

In short the value of the expression (which is known at compile time, because it is a constant expression) is representable in the type of the variable that is byte.

考虑你的表情

 final byte x = 1;
 final byte y = 2;
 byte z = x + y;//This is constant expression and value is known at compile time

因此，当求和适合字节时，它不会引发编译错误.

So as summation fits into byte it does not raise an compilation error.

现在如果你这样做

final byte x = 100;
final byte y = 100;
byte z = x + y;// Compilation error it no longer fits in byte

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