我有以下代码:
public class Tests {
public static void main(String[] args) throws Exception {
int x = 0;
while(x<3) {
x = x++;
System.out.println(x);
}
}
}
我们知道他应该只写x++或x=x+1，但在x=x+ +时，它应该先把x归为自己，然后再加1。为什么x的值是0 ?
- - -更新
下面是字节码:
public class Tests extends java.lang.Object{
public Tests();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]) throws java.lang.Exception;
Code:
0: iconst_0
1: istore_1
2: iload_1
3: iconst_3
4: if_icmpge 22
7: iload_1
8: iinc 1, 1
11: istore_1
12: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream;
15: iload_1
16: invokevirtual #3; //Method java/io/PrintStream.println:(I)V
19: goto 2
22: return
}
我会仔细阅读说明，试着理解……
你有效地得到了以下行为。
获取x的值(它是0)作为右边的“结果”
增加x的值(x现在是1)
将右边的结果(保存为0)分配给x (x现在是0)
其思想是，后增量操作符(x++)在返回变量的值后对该变量进行增量，以便在使用该变量的方程中使用。
编辑:由于评论的原因，稍微增加了一点。像下面这样考虑。
x = 1; // x == 1
x = x++ * 5;
// First, the right hand side of the equation is evaluated.
==> x = 1 * 5;
// x == 2 at this point, as it "gave" the equation its value of 1
// and then gets incremented by 1 to 2.
==> x = 5;
// And then that RightHandSide value is assigned to
// the LeftHandSide variable, leaving x with the value of 5.
这是因为x的值根本没有增加。
x = x++;
等于
int temp = x;
x++;
x = temp;
解释:
让我们看看这个操作的字节代码。考虑一个示例类:
class test {
public static void main(String[] args) {
int i=0;
i=i++;
}
}
现在运行类反汇编程序，我们得到:
$ javap -c test
Compiled from "test.java"
class test extends java.lang.Object{
test();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]);
Code:
0: iconst_0
1: istore_1
2: iload_1
3: iinc 1, 1
6: istore_1
7: return
}
现在Java虚拟机是基于堆栈的，这意味着对于每个操作，数据将被推入堆栈，从堆栈中，数据将弹出来执行操作。还有另一种数据结构，通常是存储局部变量的数组。局部变量的id是数组的索引。
让我们看看main()方法中的助记符:
iconst_0: The constant value 0
is pushed on to the stack.
istore_1: The top element of the
stack is popped out and stored in the
local variable with index 1 which
is x.
iload_1 : The value at the
location 1 that is the value of x
which is 0, is pushed into the stack.
iinc 1, 1 : The value at the
memory location 1 is incremented by 1. So x now becomes
1.
istore_1 : The value at the top of
the stack is stored to the memory location1. That is 0 is assigned
to x overwriting its incremented value.
因此，x的值不会改变，从而导致无限循环。
