用Java长时间移位 [英] Bitshifting a long in Java

问题描述

对于任何人来说，这都是一件容易的事！

Im sure this is an easy one for whoever sees it first!

为什么在Java中代码类似

Why in Java does code like

   long one = 1 << 0;
   long thirty = 1 << 30;
   long thirtyOne = 1 << 31;
   long thirtyTwo = 1 << 32;

   System.out.println(one+" = "+Long.toBinaryString(1 << 0));
   System.out.println(thirty+" = "+Long.toBinaryString(1 << 30));
   System.out.println(thirtyOne+" = "+Long.toBinaryString(1 << 31));
   System.out.println(thirtyTwo+" = "+Long.toBinaryString(1 << 32));

打印

1 = 1
1073741824 = 1000000000000000000000000000000
-2147483648 = 1111111111111111111111111111111110000000000000000000000000000000
1 = 1

这对我来说没有意义. long是一个64位数字-而在上面它的作用类似于int.我知道移位的byte会进行int提升，但是在这种情况下我看不到发生了什么.

This does not make sense to me. long is a 64 bit number - whereas it seems to act like an int in the above. I know bitshifted bytes undergo int promotion but I dont see whats going on in this case.

任何有关这里发生的事情的指针都将是不错的:)

Any pointers on whats going on here would be good :)

Thx

编辑:感谢您提供所有答案-单击提交"后，我意识到发生了什么，但是我进入了readonly模式，因此无法删除！非常感谢！

EDIT: thanks for all the answers - i realised what was going on as soon as I clicked 'submit' but SO went into readonly mode and I couldnt delete! Many thanks!

推荐答案

这是因为1是int文字，因此<<应用于整数.结果被强制转换为long，但那时为时已晚.

It's because 1 is an int literal, so << is applied to an integer. The result is cast to a long, but by then it's too late.

如果您写1L<< 32，等，那么一切都会好起来的. L用于表示long文字.

If you write 1L << 32, etc., then all will be well. L is used to denote a long literal.

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