指针和数组。 [英] Pointers and arrays.
问题描述
#include< string.h>
#include< stdio.h>
#include< windows.h>
int main（int argc，char * argv []）
{
char s [10] =" welcome";
char * p = s;
char ** prt = p; //错误发生在这里
返回0;
}
#include <string.h>
#include <stdio.h>
#include <windows.h>
int main(int argc, char *argv[])
{
char s[10] = "welcome";
char *p = s;
char **prt = p; //error happens here
return 0;
}
推荐答案
" raashid bhatt" < ra ********** @ gmail.comwrote in message
news：7c ******************** ************** @ t1g2000p ra.googlegroups.com ...
"raashid bhatt" <ra**********@gmail.comwrote in message
news:7c**********************************@t1g2000p ra.googlegroups.com...
#include< string.h>
#include< stdio.h>
#include< windows.h>
#include <string.h>
#include <stdio.h>
#include <windows.h>
< windows.his不是标准C头。请
省略此处发布的代码（无论如何都没有
与你的问题有关）。
<windows.his not a standard C header. Please
omit such from code posted here (it has nothing
to do with your question anyway).
>
int main（int argc，char * argv []）
{
char s [10] =" welcome";
char * p = s;
char ** prt = p; //错误发生在这里
>
int main(int argc, char *argv[])
{
char s[10] = "welcome";
char *p = s;
char **prt = p; //error happens here
那是因为你试图分配
一个不兼容的类型。写道：
char ** prt =& p;
That''s because you''re trying to assign
an incompatible type. Write:
char **prt = &p;
返回0;
}
return 0;
}
我已经向您展示了正确的语法。但是你想要做什么呢？b
$ b -Mike
I''ve shown you correct syntax. But what
exactly are you trying to do?
-Mike
8月19日， 6：54 * am，Mike Wahler < mkwah ... @ mkwahler.netwrote：
On Aug 19, 6:54*am, "Mike Wahler" <mkwah...@mkwahler.netwrote:
" raashid bhatt" < raashidbh ... @ gmail.com写信息
新闻：7c *********************** *********** @ t1g2000p ra.googlegroups.com ...
"raashid bhatt" <raashidbh...@gmail.comwrote in message
news:7c**********************************@t1g2000p ra.googlegroups.com...
#include< string.h>
#include< stdio.h>
#include< windows.h>
#include <string.h>
#include <stdio.h>
#include <windows.h>
< windows.his不是标准C头。请
省略此处发布的代码（无论如何都没有
与你的问题有关）。
<windows.his not a standard C header. Please
omit such from code posted here (it has nothing
to do with your question anyway).
int main（int argc，char * argv []）
{
int main(int argc, char *argv[])
{
char s [10] ="欢迎英寸;
char s[10] = "welcome";
char * p = s;
char ** prt = p; //错误发生在这里
char *p = s;
char **prt = p; //error happens here
那是因为你试图分配
一个不兼容的类型。 *写：
char ** prt =& p;
That''s because you''re trying to assign
an incompatible type. *Write:
char **prt = &p;
返回0;
}
return 0;
}
我已经向你展示了正确的语法。 *但是你想要做什么
？
-Mike
I''ve shown you correct syntax. *But what
exactly are you trying to do?
-Mike
这是错误的人为什么我需要& p作为p包含地址
this is wrong man why do i need &p as p consists the address
8月19日，15：01，raashid bhatt< raashidbh ... @ gmail.comwrote：
On 19 Aug, 15:01, raashid bhatt <raashidbh...@gmail.comwrote:
8月19日，6：54 * am，Mike Wahler < mkwah ... @ mkwahler.netwrote：
On Aug 19, 6:54*am, "Mike Wahler" <mkwah...@mkwahler.netwrote:
" raashid bhatt" < raashidbh ... @ gmail.comwrote in message
"raashid bhatt" <raashidbh...@gmail.comwrote in message
news：7c *************** ******************* @ t1g2000p ra.googlegroups.com ...
news:7c**********************************@t1g2000p ra.googlegroups.com...
#include< string.h>
#include< stdio.h>
#include< windows.h>
#include <string.h>
#include <stdio.h>
#include <windows.h>
< windows.his不是标准C头。请
省略此处发布的代码（无论如何都没有
来处理您的问题）。
<windows.his not a standard C header. Please
omit such from code posted here (it has nothing
to do with your question anyway).
int main（int argc，char * argv []）
{
int main(int argc, char *argv[])
{
char s [10] =" welcome" ;;
char s[10] = "welcome";
char * p = s;
char ** prt = p; //错误发生在这里
char *p = s;
char **prt = p; //error happens here
那是因为你试图分配
an不兼容的类型。 *写：
That''s because you''re trying to assign
an incompatible type. *Write:
char ** prt =& p;
char **prt = &p;
return 0;
}
return 0;
}
我已经向您展示了正确的语法。 *但是你确实想要做什么？
？
I''ve shown you correct syntax. *But what
exactly are you trying to do?
-Mike
-Mike
这是错误的人为什么我需要& p因为p包含地址
this is wrong man why do i need &p as p consists the address
没有它*不是*错误。考虑类型，p的类型为char *（ptr-to-
char）
prt的类型为char **（ptr-to -ptr-to-char）。它们不一样。
但是& p是ptr-to-char的地址所以它的类型为ptr-to-ptr-to-
char。
所以这项任务还可以。
获得一本好的教科书。并阅读它。
-
Nick Keighley
no it *isn''t* wrong. Consider the types, p has type char* (ptr-to-
char)
prt has type char** (ptr-to -ptr-to-char). They are not the same.
But &p is the address of a ptr-to-char so it has type ptr-to-ptr-to-
char.
So that assignment is ok.
Get a good text book. And read it.
--
Nick Keighley
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