请告诉它有效 [英] please tell it works
问题描述
hai to everybody
i我正在尝试一些编码，因为这样的流程是否有效
代码是
int * p;
* p = 20
printf（"％d"，* p）;
hai to everybody
i am trying some coding as such flow it works or not
code is
int *p;
*p=20
printf("%d",*p);
推荐答案
p将包含一些垃圾地址使此代码极易受到攻击
分段错误:-(
-
GJ
" venkatesh"< pv *********** @ gmail.com>写在留言中
news：11 ** ******************** @ g47g2000cwa.googlegr oups.com ...
p will be containing some garbage address make this code highly vulnerable
for segmentation fault :-(
--
GJ
"venkatesh" <pv***********@gmail.com> wrote in message
news:11**********************@g47g2000cwa.googlegr oups.com...
hai to everyone's
我正在尝试一些编码这样的流程是否有效
代码是
int * p;
* p = 20
printf（"％d"，* p）;
hai to everybody
i am trying some coding as such flow it works or not
code is
int *p;
*p=20
printf("%d",*p);
----- BEGIN PGP签名消息-----
哈希：SHA1
venkatesh写道：
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
venkatesh wrote:
hai to everyonebody
我正在尝试一些编码，因为这样的流程是否有效
代码是
int * p;
* p = 20
printf（"％d"，* p）;
hai to everybody
i am trying some coding as such flow it works or not
code is
int *p;
*p=20
printf("%d",*p);
除了事实上这些陈述不构成有效的C程序或
函数，因此它们是不可编译的，我只看到你的代码有一个问题。
。您将p声明为/指针/指向整数，但从不将p初始化为
包含指针值。
int * p;
单独
不足以允许正确操作
* p = 20;
因为p没有值。
你的代码看起来应该是......
#include< stdio.h> ; / * for printf（）prototype * /
void myfunction（void）
{
int * p; / *指针的位置* /
int q; / *一个整数的位置* /
p =& q; / *初始化p指向int q * /
* p = 20; / *初始化p现在指向* /
printf（"％d \ n"，* p）; / *打印int值* /
}
- -
Lew Pitcher
IT专家，企业数据系统，
企业技术解决方案，道明银行金融集团
（表达的意见是我自己的，而不是我的雇主'）
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= RpKA
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Besides the fact these statements do not constitute a valid C program or
function, and thus are uncompilable as they stand, I only see one problem with
your code. You declare p as a /pointer/ to an integer, but never initialize p to
contain a pointer value. the
int *p;
alone is insufficient to permit the proper operation of
*p=20;
because p does not have a value.
Your code should look something like....
#include <stdio.h> /* for printf() prototype */
void myfunction(void)
{
int *p; /* a place for a pointer */
int q; /* a place for an integer */
p = &q; /* initialize p to point to int q */
*p = 20; /* initialize the int that p now points to */
printf("%d\n",*p); /* print the int value */
}
- --
Lew Pitcher
IT Specialist, Enterprise Data Systems,
Enterprise Technology Solutions, TD Bank Financial Group
(Opinions expressed are my own, not my employers'')
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< br>
venkatesh写道：
venkatesh wrote:
hai to everyone
我正在尝试一些编码，因为这样的流程是否有效
代码是
int * p;
* p = 20
printf（"％d"，* p）;
hai to everybody
i am trying some coding as such flow it works or not
code is
int *p;
*p=20
printf("%d",*p);
假设你通过将
放在一个函数中并使用相应的include来使这段代码在语法上合法化：
#include< stdio.h>
无效xample（void）
{
int * p;
* p = 20;
printf（" ％d"，* p）;
}
然后人们可以合理地问`* p = 20`，" where / do / you
认为20将被存储？
还可以希望进一步的，换行终止，
输出应该在`example`被调用之后发生。
-
克里斯电刺猬 Dollin
Il Principe - Byzantium - Hansa - Antike - King''s Progress
Farfalia - Mu - Havoc - Tigris&幼发拉底河[kartenspiel]
Suppose you made this code syntactically legal by putting it
inside a function and with the appropriate include:
#include <stdio.h>
void example(void)
{
int *p;
*p = 20;
printf( "%d", *p );
}
Then one could reasonably ask of `*p = 20`, "where /do/ you
think the 20 will be stored?".
One could also hope that some further, newline-terminated,
output is supposed to happen after `example` gets called.
--
Chris "electric hedgehog" Dollin
Il Principe - Byzantium - Hansa - Antike - King''s Progress
Farfalia - Mu - Havoc - Tigris & Euphrates [kartenspiel]
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