我可以在分配的内存上做任何事吗？ [英] Can I do ANYTHING on allocated memory?
问题描述
以下任何一种情况是否构成未定义的行为？一旦我分配内存，
我可以用它做任何我想做的事情（例如，使用内存分配为
char *，作为我存储整数的内存）？
#include< stdio.h>
#include< stdlib.h>
int main （无效）{
char * a;
int * x，* y;
a =（char *）malloc（2 * sizeof（ int））;
//这里是'我想知道的部分
x =（int *）a;
y = （int *）a + sizeof（int）;
* x = 10;
* y = 20;
printf（"％ d％d \ n"，* x，* y）;
返回0;
}
-
世界上很容易接受世界的观点;孤独轻松
住在我们自己之后;但伟大的人就是在人群中间的人。
保持完美的甜蜜独处的孤独。
Ralph Waldo Emerson，自力更生1841年/> http://pinpoint.wordpress.com/
Does any of the following constitute undefined behavior? Once I allocate memory,
can I do with it whatever I want (as in the example, use memory allocated as
char *, as memory where I store integers)?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char *a;
int *x, *y;
a = (char *) malloc(2 * sizeof(int));
// here''s the part I''m wondering about
x = (int *) a;
y = (int *) a + sizeof(int);
*x = 10;
*y = 20;
printf ("%d %d\n", *x, *y);
return 0;
}
--
"It is easy in the world to live after the world''s opinion; it easy in solitude
to live after our own; but the great man is he who in the midst of the crowd
keeps with perfect sweetness the independence of solitude."
Ralph Waldo Emerson, Self-reliance 1841
http://pinpoint.wordpress.com/
推荐答案
" Sourcerer" < en **** @ MAKNIgmail.com在消息中写了
"Sourcerer" <en****@MAKNIgmail.comwrote in message
以下任何一种情况是否构成未定义的行为？一旦我分配了
内存，我可以随心所欲地使用它（例如，使用内存
分配为char *，作为存储整数的内存）？
#include< stdio.h>
#include< stdlib.h>
int main （无效）{
char * a;
int * x，* y;
a =（char *）malloc（2 * sizeof（ int））;
//这里是'我想知道的部分
x =（int *）a;
y = （int *）a + sizeof（int）;
* x = 10;
* y = 20;
printf（"％ d％d \ n"，* x，* y）;
返回0;
}
Does any of the following constitute undefined behavior? Once I allocate
memory, can I do with it whatever I want (as in the example, use memory
allocated as char *, as memory where I store integers)?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char *a;
int *x, *y;
a = (char *) malloc(2 * sizeof(int));
// here''s the part I''m wondering about
x = (int *) a;
y = (int *) a + sizeof(int);
*x = 10;
*y = 20;
printf ("%d %d\n", *x, *y);
return 0;
}
你正在使用C型系统快速和松散地玩耍。
它可以帮助你避免对齐问题和其他恶意，但是
用于向后兼容和某些低 - 等级操作，C还给你
你可以自由地搞砸它。
技术上我认为你没关系因为指向char可以转换为
任何其他类型。这是从没有空格的时候的宿醉，因此使用了char * s代替。
但是你投了一个短*，然后回来，它本来是
实施定义的行为。在绝大多数机器上int *和
short * s是相同的，而演员阵容是无操作。然而，只有可能
，short *会抛出一些int *需要的东西，所以符合要求的b $ b实现可以做任何它想要的响应。
You are playing fast and loose with the C type system.
It is there to help you avoid alignment problems and other nasties, however
for backwards compatibility and certain low-level operations, C also gives
you the freedom to mess about with it.
Technically I think you are OK because pointers to char can be converted to
any other type. This is a hangover from the days when there were no void
pointers, so char *s were used instead.
However had you cast to a short *, and back, it would have been
implentation-defined behaviour. On the vast majority of machines int * and
short *s are the same and the cast is a no op. However it is just possible
that the short * throws some bits away which an int * needs, so a conforming
implementation could do anything it wants in response.
在文章< eo ********** @ ss408.t-com.hr>中，
Sourcerer< en * ***@MAKNIgmail.com写道：
In article <eo**********@ss408.t-com.hr>,
Sourcerer <en****@MAKNIgmail.comwrote:
>一旦我分配了内存，我可以随心所欲地使用它（如示例所示，使用内存分配为
char *，作为存储整数的内存）？
>Once I allocate memory,
can I do with it whatever I want (as in the example, use memory allocated as
char *, as memory where I store integers)?
你没有把它分配为char *：
You didn''t allocate it as char *:
a =（char *）malloc（ 2 * sizeof（int））;
a = (char *) malloc(2 * sizeof(int));
您刚刚分配了一些内存，并且（不必要地）将其转换为char *
，然后再将其分配给。 malloc（）返回的内存并不是
有类型。
你可以存储你喜欢的任何东西，只要它适合你和你
尊重对齐规则。
You just allocated some memory, and (unnecessarily) cast it to char *
before assigning it to a. The memory returned by malloc() doesn''t
have a type.
You can store whatever you like in it, provided that it fits and you
respect the alignment rules.
x =（int *）a;
x = (int *) a;
罚款。
但你在这里有一个错误：
Fine.
But you have a mistake here:
y =（int *）a + sizeof（int）;
y = (int *) a + sizeof(int);
您已将a转换为整数指针，然后将sizeof（int）添加到
。如果sizeof（int）是4，那么你刚刚添加了四个int大小，
即16字节。添加指针会增加
指向对象大小的单位，所以你想要
y =（int *）（a + sizeof（ int））;
或
y =（int *）a + 1;
- Richard
-
在一些字母表中需要考虑多达32个字符
- 1963年的X3.4。
You have converted a to an integer pointer, then added sizeof(int) to
it. If sizeof(int) is 4, you have just added four int sizes to it,
i.e. 16 bytes. Adding to a pointer adds in units of the size of the
pointed-to object, so you want
y = (int *) (a + sizeof(int));
or
y = (int *) a + 1;
-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
文章< lK ********************* @ bt .com>，
Malcolm McLean< re ******* @ btinternet.comwrote：
In article <lK*********************@bt.com>,
Malcolm McLean <re*******@btinternet.comwrote:
>你在玩使用C型系统快速而松散。
>You are playing fast and loose with the C type system.
不，除了他添加的错误，他正在使用指针
正确。
No, apart from his mistake with the addition, he is using pointers
correctly.
>技术上我认为你没事，因为指向char的指针可以转换为
任何其他类型。这是从没有无效指针的日子开始的宿醉，所以使用char * s代替。
>Technically I think you are OK because pointers to char can be converted to
any other type. This is a hangover from the days when there were no void
pointers, so char *s were used instead.
void *已将char *替换为通用指针类型，但你不能在$ * b b上对void *指针进行算术运算。 char *（或unsigned char *）是用于以字节为单位访问内存的
类型，它是C存储的单位。
C保证无效*和char *具有相同的表示形式，但
OP不依赖于此。
- Richard
- -
考虑在一些字母表中需要多达32个字符
- 1963年的X3.4。
void * has replaced char * as the generic pointer type, but you can''t
do arithmetic on void * pointers. char * (or unsigned char *) is the
type used to access memory as bytes, which are the units of C storage.
C guarantees that void * and char * have the same representation, but
the OP is not relying on that.
-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
这篇关于我可以在分配的内存上做任何事吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
