java 赋值的左侧必须是变量
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The left-hand side of an assignment must be a variable
提问by Daniel
Why doesn't this work?
为什么这不起作用？
private List<Integer> xShot = new ArrayList<Integer>();
...codes
...codes
...codes
...codes
xShot.get(0) += 5;
Can't understand why the left-hand side of an assignment′isn't is a variable..
无法理解为什么赋值的左侧不是一个变量..
Someone help?
有人帮忙吗？
采纳答案by Philipp Reichart
If you just want to increment by 5 and aren't limited to List<Integer>specifically, you could avoid arguably verbose xShot.set(0, xShot.get(0) + 5)and do this instead:
如果你只想增加 5 并且不限于List<Integer>特定的，你可以避免冗长的说法xShot.set(0, xShot.get(0) + 5)，而是这样做：
List<AtomicInteger> xShot = new ArrayList<AtomicInteger>();
xShot.get(0).addAndGet(5);
This will increment the value of the AtomicIntegerin xShot.get(0)by 5 in-place without further ado.
这将使AtomicIntegerin的值xShot.get(0)就地增加 5，无需多说。
回答by Ryan Stewart
xShot.get(0)is a method call that returns a value. A variable is something you declare with a type that holds a value, like int x;, String name;, or List<Integer> xShotfrom your example. Those are the only things in Java that you can assign a value to using an assignment operator.
xShot.get(0)是一个返回值的方法调用。变量是你保存的值，喜欢的类型声明的东西int x;，String name;或者List<Integer> xShot从你的榜样。这些是 Java 中唯一可以使用赋值运算符赋值的东西。
回答by Ryan Amos
Although xShot.get(0) is a number, it is not a variable. You need to provide a variable for this to work. That said
尽管 xShot.get(0) 是一个数字，但它不是一个变量。您需要提供一个变量才能使其工作。那说
int i = xShot.get(0);
i += 5;
Will not work. iwill be incremented by 5, but xShot's object in location 5 is not the same object. You need to get, modify, and set the variable.
不管用。i将增加 5，但 xShot 在位置 5 的对象不是同一个对象。您需要获取、修改和设置变量。
For example:
例如：
xShot.set(0, xShot.get(0) + 5);
回答by Bohemian
xShot.get(0)returnsan object; it isn't a variable, so you can't assign to it.
xShot.get(0)返回一个对象；它不是一个变量，所以你不能分配给它。
Also, Integeris immutable (you can't change its value), so you would have to replacethe object at position 0with a newIntegerthat has the calculated value.
此外，Integer是不可变的（您不能更改其值），因此您必须用具有计算值的新对象替换位置处的对象。0Integer
You can achieve the intention of that line like this:
您可以像这样实现该行的意图：
xShot.set(0, xShot.get(0) + 5);
回答by Martijn Courteaux
It is like saying in Java:
这就像在 Java 中说的：
5 = 6; // "Assign 5 to 6"
The left side (5) isn't variable.
左侧 ( 5) 不是可变的。
Why is this example statement relevant? Because of Java uses always"pass by value". Which means that the return valueof a method is also "return by value".
This is pure mathematical: you can't change a value, you can change a variable. The same for Java. Five can never become six.
In other words: Only a value can be assigned to a variable.
为什么这个示例语句是相关的？因为 Java 使用总是“按值传递”。这意味着方法的返回值也是“按值返回”。这是纯粹的数学：你不能改变一个值，你可以改变一个变量。对于 Java 也是如此。五永远不可能变成六。
换句话说：只能为变量赋值。
So, the correct way of doing what you want is:
所以，做你想做的正确方法是：
xShot.set(0, xShot.get(0) + 5);
Edit:In your situation: xShot.get(int)doesn't return a variable, but a value.
编辑：在您的情况下：xShot.get(int)不返回变量，而是返回值。
回答by Bhaskar
The left-hand side of the assignment has to be explicitly a variable because in your case of statement , the expression could be a constant also , which would be a error if allowed
赋值的左侧必须明确是一个变量，因为在你的语句中，表达式也可以是一个常量，如果允许，这将是一个错误
