问题描述
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char1是一个int
cis achar1is anint
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 tochar, so its value is compatible with the variable's type;c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.c++is of typechar, so no cast is required.
这篇关于在java中如果“char c = 'a'"为什么“c = c + 1"?不编译?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！