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……
从http://download.oracle.com/javase/tutorial/java/nutsandbolts/op1.html
The increment/decrement operators can
be applied before (prefix) or after
(postfix) the operand. The code
result++; and ++result; will both end
in result being incremented by one.
The only difference is that the prefix
version (++result) evaluates to the
incremented value, whereas the
postfix version (result++) evaluates
to the original value. If you are
just performing a simple
increment/decrement, it doesn't really
matter which version you choose. But
if you use this operator in part of a
larger expression, the one that you
choose may make a significant
difference.
为了说明这一点，请尝试以下方法:
int x = 0;
int y = 0;
y = x++;
System.out.println(x);
System.out.println(y);
它会输出1和0。
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.
X = x++的工作方式如下:
首先，它计算表达式x++。对该表达式求值产生一个表达式值(即x在递增前的值)，并使x递增。
之后，它将表达式值赋给x，覆盖增量值。
因此，事件的序列如下所示(这是一个实际的反编译字节码，由javap -c生成，带有我的注释):
8: iload_1 // Remember current value of x in the stack
9: iinc 1, 1 // Increment x (doesn't change the stack)
12: istore_1 // Write remebered value from the stack to x
作为比较，x = ++x:
8: iinc 1, 1 // Increment x
11: iload_1 // Push value of x onto stack
12: istore_1 // Pop value from the stack to x
