[英]Why can't the left hand side of an assignment be an increment expression?
Could any one please tell me the meaning of "++" with array in the following code in Java: 有没有人请告诉我Java中以下代码中带有数组的“++”的含义:
int [ ] arr = new int[ 4 ];
for(int i = 0; i < arr.length; i++){
arr[ i ] = i + 1;
System.out.println(arr[ i ]++);
}
what is arr[ i ]++
meaning in above code, and why we can't do like: 什么是
arr[ i ]++
含义在上面的代码中,为什么我们不能这样做:
arr[ i ]++ = i + 1;
The operator being discussed here is called the postfix increment operator ( JLS 15.14.2 ). 这里讨论的运算符称为后缀增量运算符( JLS 15.14.2 )。 It is specified to behave as follows:
它被指定为表现如下:
- At run time, if evaluation of the operand expression completes abruptly, then the postfix increment expression completes abruptly for the same reason and no incrementation occurs.
在运行时,如果操作数表达式的评估突然完成,则后缀增量表达式出于同样的原因突然完成,并且不会发生增量。
- Otherwise, the value 1 is added to the value of the variable and the sum is stored back into the variable.
否则,将值1添加到变量的值中,并将总和存储回变量中。
- Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable.
在添加之前,对值1和变量的值执行二进制数字提升(第5.6.2节)。
- If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
如果必要,在存储之前,通过缩小的基元转换(第5.1.3节)和/或经过装箱转换(第5.1.7节)将总和缩小到变量的类型。
- The value of the postfix increment expression is the value of the variable before the new value is stored .
后缀增量表达式的值是变量的值被存储在新的值之前 。
The last point is the key for this question: the reason why you can't do arr[i]++ = v;
最后一点是这个问题的关键:你不能做的原因
arr[i]++ = v;
is the same exact reason why you can't do x++ = v;
与你不能做
x++ = v;
原因完全相同x++ = v;
; ; the postfix increment expression returns a value , not a variable .
后缀增量表达式返回一个值 ,而不是一个变量 。
From JLS 15.1 Evaluation, Denotation and Result : 从JLS 15.1评估,表示和结果 :
When an expression in a program is evaluated (executed), the result denotes one of three things:
当评估(执行)程序中的表达式时,结果表示以下三种情况之一:
- A variable [...] (in C, this would be called an lvalue)
变量[...](在C中,这将被称为左值)
- A value [...]
一个值 [...]
- Nothing (the expression is said to be void)
没什么(表达被认为是无效的)
An assignment needs a variable on the left hand side, and a value is NOT a variable, and that's why you can't do x++ = v;
赋值需要在左侧的变量和值不是一个变量,这就是为什么你不能做
x++ = v;
. 。
From JLS 15.26 Assignment Operators : 来自JLS 15.26分配运营商 :
The result of the first operand of an assignment operator must be a variable, or a compile-time error occurs .
赋值运算符的第一个操作数的结果必须是变量,否则会发生编译时错误 。 This operand may be a named variable [...], or it may be a computed variable, as can result from a field [...] or an array access.
此操作数可以是命名变量[...],也可以是计算变量,可以来自字段[...]或数组访问。 [...]
[...]
The following snippet shows erroneous attempts to assign to a value , going from rather subtle to more obvious: 以下代码段显示了错误的分配值的尝试,从相当微妙到更明显:
int v = 42;
int x = 0;
x = v; // OKAY!
x++ = v; // illegal!
(x + 0) = v; // illegal!
(x * 1) = v; // illegal!
42 = v; // illegal!
// Error message: "The left-hand side of an assignment must be a variable"
Note that you can use the postfix increment operator somewhere on the left hand side of an assignment operator, as long as the final result is a variable. 请注意,您可以 某处使用后缀增量运算符在赋值运算符的左侧,只要最后的结果是一个变量。
int[] arr = new int[3];
int i = 0;
arr[i++] = 2;
arr[i++] = 3;
arr[i++] = 5;
System.out.println(Arrays.toString(arr)); // prints "[2, 3, 5]"
arr[i]++
means "increase the value of arr[i]
by 1" and return the old value. arr[i]++
表示“将arr[i]
的值增加1”并返回旧值。 So your code first sets arr[i]
to i+1
using arr[ i ] = i + 1;
因此您的代码的第一组
arr[i]
至i+1
使用arr[ i ] = i + 1;
. 。 It then increases it to
i + 2
using arr[ i ]++
and prints the value it had before it was increased the second time, ie i + 1
. 然后使用
arr[ i ]++
将其增加到i + 2
并打印第二次增加之前的值,即i + 1
。
You can't use arr[ i ] = arr[i] + 1;
你不能使用
arr[ i ] = arr[i] + 1;
instead because it means "increase the value of arr[i]
by 1 and return the new value". 相反,因为它意味着“将
arr[i]
的值增加1并返回新值”。
You can't do arr[ i ]++ = i + 1;
你不能做
arr[ i ]++ = i + 1;
because it doesn't make sense. 因为它没有意义。
The code System.out.println(arr[i]++)
means this: 代码
System.out.println(arr[i]++)
意味着:
int tmp = arr[i];
arr[i] = arr[i] + 1;
System.out.println(tmp);
Your second example doesn't make sense because you can't use the ++
operator on a value. 你的第二个例子没有意义,因为你不能在值上使用
++
运算符。
The ++ operator has nothing to do with arrays. ++运算符与数组无关。 It increments any integer variable (or more generally, any lvalue) by 1. It's the same as the i++ in the loop.
它将任何整数变量(或更一般地,任何左值)递增1.它与循环中的i ++相同。
You can write either ++x or x++. 您可以编写++ x或x ++。 These both increment x, but they have different values: ++x returns the new value of x and x++ returns the old value.
这些都增加x,但它们具有不同的值:++ x返回x的新值,x ++返回旧值。 Thus, your code prints 1, 2, 3, 4 instead of 2, 3, 4, 5.
因此,您的代码打印1,2,3,4而不是2,3,4,5。
arr[ i ]++
increases arr[i] by 1. It could be like: arr[ i ]++
将arr [i]增加1.它可能是这样的:
arr[i] = i + 1;
As for arr[ i ]++ = i + 1;
至于
arr[ i ]++ = i + 1;
please do not try to write such code. 请不要尝试编写此类代码。 Even if it compiles it will be puzzle for you or for others.
即使它编译它也会让你或其他人感到困惑。
PS I would prefer: PS我更喜欢:
++arr[i];
arr[i]++ is increase the value of arr[i] by one and assign arr [i] ++将arr [i]的值增加1并赋值
as for arr[ i ]++ = i + 1;
至于
arr[ i ]++ = i + 1;
This phrase means something entirely different, I don't know that it is even valid java to be honest. 这句话意味着完全不同的东西,我不知道它甚至是有效的Java诚实。 I would, if it works, incriment the value at arr[i] and then asign it to the index + 1;
如果它有效,我会在arr [i]中输入值,然后将其指向索引+ 1;
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