导致 java:240 的 switch 语句(可能尚未初始化)
我正在尝试根据输入将值存储在变量中:
I am trying to store a value inside a variable depending on the input:
switch(pepperoni) {
case 'Y':
case 'y':
topping1 = 1;
break;
case 'N':
case 'n':
topping1 = 0;
break;
default:
{
System.out.print("This is not a valid response, please try again
");
System.out.print("Do you want Pepperoni? (Y/N): ");
pepperoni = scan.next().charAt(0);
break;
}
我希望变量 topping1 如果输入为Y"或y"则存储值 1，如果输入为N"或n"则存储值 0
I want the variable topping1 to store the value 1 if the input is 'Y' or 'y' and to store the value 0 if the input is 'N' or 'n'
如果输入既不是Y"、y"、N"也不是n"，那么我希望它重复问题，直到输入有效的输入.
If the input is neither 'Y', 'y', 'N' nor 'n' then I want it to repeat the question until a valid input is typed in.
当我稍后在程序中尝试打印值因为它可能尚未初始化"时，问题就出现了，这在某种程度上是有道理的.(下例)
The problem arises when I later in the program try to print the value 'because it might have not been initialized', which somewhat makes sense. (example below)
if(topping1 > 0)
System.out.println("Pepperoni");
// 243: error: variable topping1 might not have been initialized
我确实意识到还有其他方法可以做到这一点，但由于我真的很想学习 Java，所以我尝试尽可能多地了解基础知识.因此，如果有人能告诉我为什么这不起作用以及是否有办法通过 switch 语句或快速修复来做到这一点，我会非常高兴.
I do realize there are other ways to do this, but as I am really wanting to learn Java I try to understand as much of the fundamentals as possible. Therefore would I be really happy if someone could tell me why this not work and if there is a way to do this with a switch statement or quick fixes.
推荐答案
如果 pepperoni 不是 Y, y, N 或 n，你永远不会给 topping1 赋值，因为 default 的情况永远不会给它赋值.例如，如果 pepperoni 不是这四个值之一，则控制流会跳过其他两种情况并转到 default，它永远不会给出 topping1 一个值，所以稍后当你尝试使用它时，可能 topping1 根本就没有收到过一个值.
If pepperoni is not Y, y, N, or n, you never assign a value to topping1, because the default case never assigns it a value. E.g., if pepperoni is not one of those four values, then the flow of control skips the other two cases and goes to default, which never gives topping1 a value, so later when you try to use it, it's possible topping1 has never received a value at all.
解决方法"是更正逻辑，这样您就永远不会在未为其分配值的情况下尝试使用 topping1.如何你这样做取决于你没有向我们展示的逻辑.例如，您可以为其分配一个不同于 0 或 1 的值(您在 switch 的其他分支中分配的值).
The "workaround" is to correct the logic so that you never try to use topping1 without having assigned it a value. How you do that depends on logic you haven't shown us. You might assign it a value other than 0 or 1 (the values you assign in the other branches of the switch), for instance.
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