为什么在while循环中递增int原语不会永远循环 [英] Why incrementing int primitive in while loop does not loop forever

问题描述

我有以下代码示例

int i = 1;  

while(i != 0) {  
    i++;  
} 

我期望它可以无限循环运行，但事实并非如此.然后，当我在while循环之后打印值时，得到了:

I was expecting this to run in an infinite loop but it didn't. Then when I printed the value after while loop I got:

i value is 0.  

任何人都可以让我知道发生了什么吗?

Can any one let me know what exactly is happening?

推荐答案

我期望它可以无限循环地运行，但事实并非如此.

I was expecting this to run in an infinite loop but it didn't.

不，不会. 最终该值将再次变为0.

No, it wouldn't. Eventually the value will become 0 again.

尤其是，它将在逻辑上像这样执行:

In particular, it will logically execute like this:

1
2
3
...
2,147,483,646
2,147,483,647  
-2,147,483,648  // Integer overflow => wrapping
-2,147,483,647
...
-3
-2
-1
0

...然后循环将结束

... then the loop will end

来自JLS的第15.18.2节:

如果整数加法溢出，则结果是数学和的低阶位，以某种足够大的二进制补码格式表示.如果发生溢出，则结果的符号与两个操作数值的数学和的符号不同.

If an integer addition overflows, then the result is the low-order bits of the mathematical sum as represented in some sufficiently large two's-complement format. If overflow occurs, then the sign of the result is not the same as the sign of the mathematical sum of the two operand values.

此代码没有内存问题，因为它不执行任何分配-只是在整个堆栈上递增局部变量.

There are no memory issues with this code because it doesn't perform any allocations - it just increments a local variable, entirely on the stack.

现在是否真的是 是另一回事-如果JIT能够在不执行每次迭代的情况下弄清行为将是什么，它就可以对其进行优化-但这是一个实施细节.重要的部分是了解行为背后的逻辑.

Now whether or not it actually does that is a different matter - if the JIT is able to work out what the behaviour will be without executing every iteration, it can optimize it away - but that's an implementation detail. The important part is understanding the logic behind the behaviour.

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