比特移位与架构问题。 [英] Bit shifting versus architecture question.
问题描述
我理解不同的处理器硬件可能会以不同的顺序将这些位存储在一个
字节中。它是否会在C中产生差异
作为位移无符号字符变量？
例如，如果我有
unsigned char x = 1;
总是如此
（x << 1）== 2
（x< ;< 2）== 4
等？
或者我可能要改变某些转变的方向
架构？
（我正在抓住吸管试图找出为什么在我的i86系统上编写的某些C软件
在另一个系统上编译失败的原因
个人的PPC系统。）
感谢您的帮助。
问候，
Charles Sullivan
I understand different processor hardware may store the bits in a
byte in different order. Does it make a difference in C insofar
as bit-shifting unsigned char variables is concerned?
E.g, if I have
unsigned char x = 1;
is it always true that
(x << 1) == 2
(x << 2) == 4
etc?
Or might I have to reverse the direction of the shift on some
architectures?
(I''m grasping at straws trying to figure out why some C software
written on my i86 system is failing when compiled on another
individual''s PPC system.)
Thanks for your help.
Regards,
Charles Sullivan
推荐答案
Charles Sullivan< cw ****** @ triad.rr.comwrites：
Charles Sullivan <cw******@triad.rr.comwrites:
例如，如果我有
unsigned char x = 1;
总是如此
（x << 1）== 2
（x << 2）== 4
等？
E.g, if I have
unsigned char x = 1;
is it always true that
(x << 1) == 2
(x << 2) == 4
etc?
是的，这总是正确的。 C中的移位运算符定义在数值的
项中，而不是如何表示这些值。
-
这是一个很棒的答案。
这是偏离主题的，它是不正确的，它没有回答这个问题。
- 理查德希思菲尔德
Yes, that''s always true. The shift operators in C are defined in
terms of numeric values, not how those values are represented.
--
"This is a wonderful answer.
It''s off-topic, it''s incorrect, and it doesn''t answer the question."
--Richard Heathfield
Charles Sullivan在02/07/07 14:46写道：
Charles Sullivan wrote On 02/07/07 14:46,:
我理解不同的处理器硬件可以按不同的顺序将这些位存储在一个
字节中。它是否会在C中产生差异
作为位移无符号字符变量？
例如，如果我有
unsigned char x = 1;
总是如此
（x << 1）== 2
（x< ;< 2）== 4
等？
I understand different processor hardware may store the bits in a
byte in different order. Does it make a difference in C insofar
as bit-shifting unsigned char variables is concerned?
E.g, if I have
unsigned char x = 1;
is it always true that
(x << 1) == 2
(x << 2) == 4
etc?
是。
Yes.
或者我可能需要在某些方面改变班次的方向
架构？
Or might I have to reverse the direction of the shift on some
architectures?
No.订单只有当你将两个较小的块组装成一个较大的块时才会发挥作用，或者将较大的块分解成较小的块。像大多数C运算符一样，移位运算符和b
代表的价值一样，无论它如何表示。因此，
即使这样：
int x = 128;
断言（x << 1 == 256） ;
断言（x <<<< 2 == 512）;
在所有系统上运行相同，包括Big-和LittleEndian，
，即使单独的一位（可能）从一个字节移动到另一个字节
。
No. "Order" only comes into play when you assemble two
or more smaller pieces into one larger piece, or decompose
the larger piece into smaller fragments. The shift operators,
like most C operators, works with the value that all the
pieces represent, no matter how it is represented. Thus,
even this:
int x = 128;
assert (x << 1 == 256);
assert (x << 2 == 512);
works identically on all systems, both Big- and LittleEndian,
even though the lone one-bit moves (probably) from one byte
of x to another.
（我'我正在抓住吸管试图找出为什么在我的i86系统上写的一些C软件
在另一个
个人的PPC系统上编译时失败了。）
(I''m grasping at straws trying to figure out why some C software
written on my i86 system is failing when compiled on another
individual''s PPC system.)
我不知道你所展示的释义是怎样的，b
负责。如果它被一些稍微改变的东西释义，你可以发布一个更大的东西的实际剪辑，
也许有人会发现有用的东西。
-
Er ********* @ sun。 com
Charles Sullivan写道：
Charles Sullivan wrote:
我理解不同的处理器硬件可能存储这些位a
字节的顺序不同。它是否会因为位移无符号字符变量而在C中产生差异？
I understand different processor hardware may store the bits in a
byte in different order. Does it make a difference in C insofar
as bit-shifting unsigned char variables is concerned?
编号硬件甚至不需要任何位。没有什么可以排除在三元计算机上实现C的实现。
No. The hardware is not even required to have any "bits". There''s nothing that
precludes a C implementation on, say, a ternary computer.
例如，如果我有
unsigned char x = 1;
总是如此
（x << 1）== 2
（x << 2）== 4
等？
E.g, if I have
unsigned char x = 1;
is it always true that
(x << 1) == 2
(x << 2) == 4
etc?
是的。 比特的概念在C中指的是传统二元的元素
表示法，而不是硬件位。 1在二进制文件中始终为1，而2在二进制文件中为bb始终为10。这意味着单个左移将始终将1
变为2.
Yes. The notion of "bit" in C refers to an element of the traditional binary
notation, not to the hardware bit. ''1'' is always ''1'' in binary , while ''2'' is
always ''10'' in binary. This means that a single left shift will always turn 1
into 2.
或者我可能必须改变转变的方向一些
架构？
Or might I have to reverse the direction of the shift on some
architectures?
No.
-
祝你好运，
Andrey Tarasevich
No.
--
Best regards,
Andrey Tarasevich
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