长型64位的Linux [英] Long type 64bit linux

问题描述

很简单的问题球员，但也许我只是忘了什么东西。
在64位Linux中，一个长8个字节是否正确？
如果是这样的话，我想设置的第64位，我可以做到以下几点：

 无符号长NUM = 1＆LT;＆LT; 63;
 

每当我编译这一点，但是，它给了我一个错误，说我左移比宽度多。
另外，如果我想需要很长的类型的第32位（无符号扩展），可我做的：

  NUM = NUM​​＆放大器; 0xFFFFFFFF的;
 

或者怎么样：

  NUM =（INT）（NUM）;
 

感谢您。

解决方案

  

在64位版，而长8个字节是否正确？

需要不是。依赖于编译器不是底层操作系统。勾选此一个不错的讨论。
什么决定的sizeof整数？

  

每当我编译这一点，但是，它给了我一个错误，说我
  左移位由以上的宽度

大家都已经回答了这一点。使用 1UL

  

另外，如果我想采取一个长类型的第32位（不
  符号扩展），我可以这样做：

  NUM = NUM​​＆放大器; 0xFFFFFFFF的;
或者怎么样：NUM =（INT）（NUM）;
 

NUM = NUM​​＆放大器; 0xFFFFFFFF的。这会给你低32位。但要注意，如果长您的系统上只有4个字节，那么你得到的整个号码的。即将符号扩展的一部分，如果你已经使用了长，而不是无符号长那么你不能做掉符号扩展位。例如， 1 重新psented所有那些$ P $，直接从第0位。您将如何通过屏蔽避免这些的？

NUM =（INT）（NUM）会给你低32位，但编译器可能会通过溢出异常警告，如果 NUM 不适合的 INT

Very simple questions guys, but maybe I'm just forgetting something. In 64bit linux, a long is 8bytes correct? If that's the case, and I want to set the 64th bit, I can do the following:

unsigned long num = 1<<63;

Whenever I compile this, however, it gives me an error saying that I'm left shifting by more than the width. Also, if I wanted to take the first 32bits of a long type (without sign extension), can I do:

num = num&0xFFFFFFFF;

or what about:

num = (int)(num);

Thank you.

解决方案

In 64bit linux, a long is 8bytes correct?

Need not be. Depends on the compiler than on the underlying OS. Check this for a nice discussion. What decides the sizeof an integer?

Whenever I compile this, however, it gives me an error saying that I'm left shifting by more than the width

Everyone have already answered this. Use 1UL

Also, if I wanted to take the first 32bits of a long type (without sign extension), can I do:

num = num&0xFFFFFFFF;
or what about:

num = (int)(num);

num = num&0xFFFFFFFF. This will give you the lower 32-bits. But note that if long is just 4 bytes on your system then you are getting the entire number. Coming to the sign extension part, if you've used a long and not unsigned long then you cannot do away with the sign extended bits. For example, -1 is represented as all ones, right from the 0th bit. How will you avoid these ones by masking?

num = (int)(num) will give you the lower 32-bits but compiler might through a Overflow Exception warning if num does not fit into an int

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