在Java中多长时间int转换工作？ [英] How does long to int cast work in Java?

问题描述

这个问题是不如何长应该正确地转换为int ，而是当我们错误地将它转换为int时会发生什么。

This question is not about how a long should be correctly cast to an int, but rather what happens when we incorrectly cast it to an int.

因此，请考虑此代码 -

So consider this code -

   @Test
    public void longTest()
    {
        long longNumber = Long.MAX_VALUE;
        int intNumber = (int)longNumber; // potentially unsafe cast.
        System.out.println("longNumber = "+longNumber);
        System.out.println("intNumber = "+intNumber);
    }

这将提供输出 -

longNumber = 9223372036854775807
intNumber = -1

现在假设我做出以下更改 -

Now suppose I make the following change-

long longNumber = Long.MAX_VALUE - 50;

然后我得到输出 -

I then get the output -

longNumber = 9223372036854775757
intNumber = -51

推荐答案

以下是数学：

1. 将 long 的值视为 2 ^ 64 加上该值。因此， -1 被视为2 ^ 64 - 1.（这是无符号 64位值，


2. 取结果和mod 2 ^ 32。 （这是无符号 32位值。）


3. 如果结果是> = 2 ^ 31，则减去2 ^ 32。 （这是签名的32位值，Java int 。）

1. Treat negative long values as 2^64 plus that value. So -1 is treated as 2^64 - 1. (This is the unsigned 64-bit value, and it's how the value is actually represented in binary.)
2. Take the result and mod by 2^32. (This is the unsigned 32-bit value.)
3. If the result is >= 2^31, subtract 2^32. (This is the signed 32-bit value, the Java int.)

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