为什么i = i +我给我0？ [英] Why does i = i + i give me 0?

问题描述

我有一个简单的程序：

public class Mathz {
    static int i = 1;
    public static void main(String[] args) {    
        while (true){
            i = i + i;
            System.out.println(i);
        }
    }
}

当我运行此程序时，我看到的只是 0 我输出中的 i 。我原本预计第一次会有 i = 1 + 1 ，然后是 i = 2 + 2 ，然后是 i = 4 + 4 等。

When I run this program, all I see is 0 for i in my output. I would have expected the first time round we would have i = 1 + 1, followed by i = 2 + 2, followed by i = 4 + 4 etc.

这是因为我们一试要在左侧重新声明 i ，其值将重置为 0 ？

Is this due to the fact that as soon as we try to re-declare i on the left hand-side, its value gets reset to 0?

如果有人能指出我的更精细的细节，那就太棒了。

If anyone can point me into the finer details of this that would be great.

更改 int 到 long ，它似乎是按预期打印数字。我对它达到最大32位值的速度感到惊讶！

Change the int to long and it seems to be printing numbers as expected. I'm surprised at how fast it hits the max 32-bit value!

推荐答案

问题是整数溢出造成的。

The issue is due to integer overflow.

在32位二进制补码算术中：

In 32-bit twos-complement arithmetic:

i 确实开始具有两个幂的值，但是一旦你到达2 ³⁰ ，溢出行为就开始了：

i does indeed start out having power-of-two values, but then overflow behaviors start once you get to 2³⁰:

2 ³⁰ + 2 ³⁰ = -2 ³¹

2³⁰ + 2³⁰ = -2³¹

-2 ³¹ + -2 ³¹ = 0

-2³¹ + -2³¹ = 0

... in int 算术。

...in int arithmetic.

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