在java中如果“char c = 'a'"为什么“c = c + 1"?不编译?

2022-01-12 00:00:00 char compiler-errors casting java primitive-types

我尝试编译以下代码:

public static void main(String[] args){
    for (char c = 'a'; c <='z'; c = c + 1) {
        System.out.println(c);
    }
}

当我尝试编译时，它会抛出:

When I try to compile, it throws:

错误:(5, 41) java: 不兼容的类型: 可能的有损转换int转char

Error:(5, 41) java: incompatible types: possible lossy conversion from int to char

问题是，如果我编写 c = (char)(c + 1)、c += 1 或 c++.

The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.

我检查过，当我尝试 char c = Character.MAX_VALUE + 1; 时编译器会抛出类似的错误，但我认为 'c' 的值无法传递 'char' 类型最大值在原始函数中.

I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.

推荐答案

c + 1 是一个 int，因为操作数经过 二进制数字提升:

c + 1 is an int, as the operands undergo binary numeric promotion:

  • c 是一个 char
  • 1 是一个 int
  • c is a char
  • 1 is an int

所以 c 必须扩展为 int 以使其兼容添加；并且表达式的结果是 int 类型的.

so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.

至于有效"的东西:

  • c = (char)(c + 1) 将表达式显式转换为 char，因此其值与变量的类型兼容；
  • c += 1 等价于 c = (char) ((c) + (1))，所以和上一个基本一样.
  • c++ 是类型 char，所以不需要强制转换.
  • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;
  • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.
  • c++ is of type char, so no cast is required.

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