或者 if 语句中带有 int 的操作数
我的问题是程序没有按照我的意图“他”读取代码。
我有
if (hero.getPos() == (6 | 11 | 16)) {
move = new Object[] {"Up", "Right", "Left"};
} else {
move = new Object[] {"Up", "Down", "Right", "Left"};
}
当英雄位置为6时，程序仍然转到其他位置。
这是为什么？是因为操作数吗？如果是，我应该如何更改？
My problem is that program is not reading codes as i intended "he" would.
I have
if (hero.getPos() == (6 | 11 | 16)) {
move = new Object[] {"Up", "Right", "Left"};
} else {
move = new Object[] {"Up", "Down", "Right", "Left"};
}
When hero position is 6, the program still goes to else.
Why is that? Is it because of operands? If yes, how should i change it?
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评论（12）
使用：
这将实现您想要的功能。
您所做的是将
hero.getPos()与(6|11|16)的结果进行比较，这将执行 这些数字之间的按位或。Use:
This will do what you want.
What you did is comparing
hero.getPos()with the result of(6|11|16)which will do bitwise or between those numbers.其他答案是正确的，只是以不同的方式思考你可以使用集合。
The other answers are correct, just thinking differently you may use Sets.
你不能那样做。它按位或 3 的数字。
你必须这样做：
你看到区别了吗？
|是按位或，而||是逻辑或。另请注意，您每次都必须重写比较。
You cannot do it like that. It ors the 3 number bitwise.
You have to do like this :
You see the difference ?
|is a bitwise or while||is a logical or.Note also that you have to rewrite the comparison each time.
(6 | 11 | 16)将首先计算为 31（二元运算），即6 != 31。不是你想要的。更好的是检查每个位置（你只有 3 个位置，所以内联很好，更多考虑使用循环）：
(6 | 11 | 16)would be evaluated first to 31 (binary operation), which is6 != 31. Not what you want.Better is to check every single position (you have only 3, so inline is good, for more consider using a loop):
不，您需要检查每个值的
ci.getNumber() == ...，或者将它们添加到集合中并检查myCollection.contains(ci.getNumber( ））。但是，如果您要根据多个已知值检查方法，您可能需要重新考虑代码的结构。No, you're going to need to check
ci.getNumber() == ...for each value, or add them to a collection and checkmyCollection.contains(ci.getNumber()). However, you may want to re-think the structure of your code if you are checking a method against several known values.使用以下答案：
如何我可以测试一个数组是否包含某个值吗？
您可以创建一个数字数组并检查您的 ci.getNumber() 是否在其中。
using the answer from:
How can I test if an array contains a certain value?
you could create an array of numbers and check if your ci.getNumber() is in it.
不。您可以创建一个
Set一次，然后使用它，或者只是：我还会考虑将这些数字更改为常量，以便获得更有意义的代码。
No. You could create a
Set<Integer>once and then use that, or just:I'd also consider changing those numbers into constants so that you get more meaningful code.
没有这样的运算符。但如果你要比较数字，你可以使用 switch 来模拟。方法如下：
希望这会有所帮助。
There is no such operator. But if you are comparing number, you can use switch do simulate that. Here is how:
Hope this helps.
我必须承认我没有检查过这一点，但逻辑看起来是正确的。
I must admit I haven't checked this but the logic looks correct.
您可以将所有数字放入一个集合中，然后使用
contains()方法。除此之外，我不认为有任何特殊的语法可以像您想要的那样进行比较。You could put all the numbers in a collection, and then use the
contains()method. Other than that, I don't believe there is any special syntax for comparing like you want to do.Java 不会让你这样做。你可以做一个哈希查找（这对于这个来说是多余的）或一个 case 语句，或者一个大的丑陋的多重比较：
Java won't let you do that. You can do a hash lookup (which is overkill for this) or a case statement, or a big honking ugly multiple compare:
不..你必须单独比较它们。
no.. you have to compare them individually.