设置并重置地址的最低有效位 [英] set and reset the least significant bit of a address
问题描述
我想设置并重置一个地址的最低有效位（
指针指向的位置）。
我试过这个，但是不正确：
#define BIT 0x1
void foo（）{
void * p;
* p = * p& ~BIT
}
I want to set and reset the least significant bit of a address (where a
pointers points to).
I tried this, but it is not correct:
#define BIT 0x1
void foo(){
void *p;
*p = *p & ~BIT
}
推荐答案
onsbomma< on ****** @ hotmail.com>潦草地写道：
onsbomma <on******@hotmail.com> scribbled the following:
我想设置并重置一个地址的最低位（指针指向的位置）。
我试过这个，但是不正确：
#define BIT 0x1
void foo（）{
void * p;
* p = * p& ~BIT
}
I want to set and reset the least significant bit of a address (where a
pointers points to).
I tried this, but it is not correct: #define BIT 0x1 void foo(){
void *p;
*p = *p & ~BIT
}
其实，& ~BIT确实是重置LSB的正确方法。 | BIT是
设置它的正确方法。你做错了是用p作为void
指针。 void指针只是一个指针 - 它没有任何
信息，它指向的是什么。您必须使用指向
整数类型的指针。哦，并确保它实际上指向一个
可修改的地址。
-
/ - Joona Palaste（pa*****@cc.helsinki.fi）-------------芬兰-------- \
\- -------------------------------------------------- -----规则！ -------- /
"根据奥卡姆的牙刷，我们只需要优化最常用的
指令。
- Teemu Kerola
Actually, & ~BIT is indeed the right way to reset the LSB. | BIT is the
right way to set it. What you''re doing wrong is using p as a void
pointer. A void pointer is only a pointer - it doesn''t have any
information of what it''s pointing to. You have to use a pointer to an
integer type instead. Oh, and make sure it''s actually pointing at a
modifiable address.
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"And according to Occam''s Toothbrush, we only need to optimise the most frequent
instructions."
- Teemu Kerola
onsbomma写道：
onsbomma wrote:
我想设置并重置一个地址的最低有效位（其中一个指针指向）。
我试过这个，但是不正确：
#define BIT 0x1
void foo（）{
void * p;
* p = * p& ~BIT
}
I want to set and reset the least significant bit of a address (where a
pointers points to).
I tried this, but it is not correct:
#define BIT 0x1
void foo(){
void *p;
*p = *p & ~BIT
}
1）
如果你想改变指针的值而不是值
指向的对象请勿使用*。
2）将指针值转换为无符号整数
存储指针。有一个预定义的类型：intptr_t。
3）
#include< stdint.h>
void * p;
intptr_t ip =（intptr_t）p; //将指针值存储在ip中
ip = ip& 〜1; //消除最不重要的位
p =（void *）ip; //再次将结果存储到指针中。
4）不要写：
p =（void *）（（intptr_t）p ）& ~1;
因为在2周内你自己会徘徊那些混乱意味着什么。写清楚。
1)
If you want to change the value of the pointer and not the value
of the object being pointed to DO NOT use the "*".
2) convert the pointer value into an unsigned integer capable
of storing a pointer. There is a predefined type for that: intptr_t.
3)
#include <stdint.h>
void *p;
intptr_t ip = (intptr_t)p; // Store the pointer value in ip
ip = ip & ~1; // eliminate the least significant bit
p = (void *)ip; // store the result into the pointer again.
4) Do not write:
p = (void *)((intptr_t)p)&~1;
because in 2 weeks you yourself will be wandering what
that mess means. Write clearly.
> onsbomma写道：
> onsbomma wrote:
。>我想设置并重置地址的最低位
。> （指针指向）。
。>我试过这个，但这不正确：
。>
。> #define BIT 0x1
。>
。> void foo（）{
。> void * p;
。> * p = * p& ~BIT
忽略p未初始化，取消引用void指针
是违反约束的。如果你想要修改一个字节，请使用指向unsigned char的指针。
jacob navia写道：1）
如果你想改变指针的值而不是指向的对象的值
不要使用*。
2）将指针值转换为无符号整数
存储指针。有一个预定义的类型：
intptr_t。
但是它只能在C99下使用，并且它是可选的
是否实现定义它。
3）
#include< stdint.h>
void * p;
intptr_t ip =（intptr_t）p; //将指针值存储在ip中
ip = ip& 〜1; //消除最不重要的位
p =（void *）ip; //再次将结果存储到指针中。
与intptr_t之间的转换是实现定义的。
这种方法可能无法做到某些（真实的）
机器。
4）不要写：
p =（void *）（（intptr_t）p）& ~1;
的确如此。 ITYM：p =（void *）（（（intptr_t）p）&〜（intptr_t）1）;
因为在2周内你自己会徘徊那些混乱意味着什么。
.> I want to set and reset the least significant bit of a address
.> (where a pointers points to).
.> I tried this, but it is not correct:
.>
.> #define BIT 0x1
.>
.> void foo(){
.> void *p;
.> *p = *p & ~BIT
Ignoring that p is not initialised, dereferencing a void pointer
is a constraint violation. Use a pointer to unsigned char if you
want to modify a byte.
jacob navia wrote: 1)
If you want to change the value of the pointer and not the value
of the object being pointed to DO NOT use the "*".
2) convert the pointer value into an unsigned integer capable
of storing a pointer. There is a predefined type for that:
intptr_t.
However it is only available under C99 and it is optional as to
whether an implementation defines it.
3)
#include <stdint.h>
void *p;
intptr_t ip = (intptr_t)p; // Store the pointer value in ip
ip = ip & ~1; // eliminate the least significant bit
p = (void *)ip; // store the result into the pointer again.
The conversion to and from intptr_t is implementation defined.
This method may well fail to do what''s expected on some (real)
machines.
4) Do not write:
p = (void *)((intptr_t)p)&~1;
Indeed. ITYM: p = (void *) ( ((intptr_t) p) & ~ (intptr_t) 1 );
because in 2 weeks you yourself will be wandering what
that mess means.
无论如何，我会很困惑地看到第3步。 ;）
跟踪p是否在奇数或
偶数字节并且在适当的时候使用 - 似乎更简单。
-
彼得
I''d be puzzled seeing step 3 anyway. ;)
It would seem simpler to keep track of whether p is at an odd or
even byte and simply use -- where appropriate.
--
Peter
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