什么是“int（* p）[3]”意思是如何使用它？ [英] What does "int (*p)[3]" mean and how to use it?
问题描述
我知道int * p [3]意味着指向int的指针数组，但我遇到了这个令人困惑的
（* p）[3]。
它是什么意思以及如何使用它？感谢您的帮助。
I know int *p[3] mean an array of pointers to int, but I meet int
(*p)[3] which is confusing.
What does it mean and how to use it? Thanks for help.
推荐答案
hn ******** @ gmail.com 写道：
我知道int * p [3]表示一个指针数组int，但我遇到了这个令人困惑的
（* p）[3]。
这意味着什么以及如何使用它？感谢帮助。
I know int *p[3] mean an array of pointers to int, but I meet int
(*p)[3] which is confusing.
What does it mean and how to use it? Thanks for help.
int（* p）[3]将p声明为指向大小为3的int数组的指针。
例如：
#include< stdio.h>
#include< stdlib.h>
int main（void）
{
int（* p）[3];
int foo [3] = {0,1， 2};
p =& foo;
printf（"％d - ％d - ％d \ n"，（* p ）[0]，（* p）[1]，（* p）[2]）;
返回EXIT_SUCCESS;
}
PS：请查看常见问题解答以获得更好的解释。
int (*p)[3] declares p as a pointer to an array of int of size 3.
example:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
int (*p)[3];
int foo[3] = {0, 1, 2};
p = &foo;
printf(" %d -- %d -- %d\n", (*p)[0], (*p)[1], (*p)[2]);
return EXIT_SUCCESS;
}
PS: Please look into the FAQ for better explanation.
是（* p）[？]等于foo [ ？]这里？
i可以使用（* p）[？]在任何地方使用foo [？]？
i意味着我可以用（* p）替换所有foo [？] ）[？]如我宣布的那样lang
"
int（* p）[3];
int foo [3] = {0,1,2};
p =& foo;
"
p _ *********** @ yahoo.co.in 写道：
is (*p)[?] equals to foo[?] here?
i can use (*p)[?] anywhere i used foo[?] ?
i mean can i replace all foo[?] with (*p)[?] as lang as i declared
"
int (*p)[3];
int foo[3] = {0, 1, 2};
p = &foo;
"
p_***********@yahoo.co.in wrote:
hn********@gmail.com 写道：
我知道int * p [3]表示指向int的指针数组，但是我遇到了这个令人困惑的
（* p）[3]。 br />
这意味着什么以及如何使用它？感谢帮助。
I know int *p[3] mean an array of pointers to int, but I meet int
(*p)[3] which is confusing.
What does it mean and how to use it? Thanks for help.
int（* p）[3]将p声明为指向大小为3的int数组的指针。
例如：
#include< stdio.h>
#include< stdlib.h>
int main（void）
{
int（* p）[3];
int foo [3] = {0,1， 2};
p =& foo;
printf（"％d - ％d - ％d \ n" ，（* p）[0]，（* p）[1]，（* p）[2]）;
返回EXIT_SUCCESS;
}
PS：请查看常见问题解答以获得更好的解释。
int (*p)[3] declares p as a pointer to an array of int of size 3.
example:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
int (*p)[3];
int foo[3] = {0, 1, 2};
p = &foo;
printf(" %d -- %d -- %d\n", (*p)[0], (*p)[1], (*p)[2]);
return EXIT_SUCCESS;
}
PS: Please look into the FAQ for better explanation.
p _ *** ********@yahoo.co.in 说：
< snip>
p_***********@yahoo.co.in said:
<snip>
>
int（* p）[3]将p声明为指向大小为3的int数组的指针。 />
例如：
#include< stdio.h>
#include< stdlib.h>
int main（无效）
{
int（* p）[3];
int foo [3] = {0,1,2};
p =& foo;
>
int (*p)[3] declares p as a pointer to an array of int of size 3.
example:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
int (*p)[3];
int foo[3] = {0, 1, 2};
p = &foo;
到目前为止一直这么好......
So far so good...
printf（"％d - ％d - - ％d \ n，（* p）[0]，（* p）[1]，（* p）[2]）;
printf(" %d -- %d -- %d\n", (*p)[0], (*p)[1], (*p)[2]);
如果使用％p，printf可以打印无效指针值，但就是这样。所以
以上的行应该是：
printf（"％p - ％p - ％p \ n"，（void *） （* p）[0]，
（void *）（* p）[1]，
（void *）（* p）[2]）; < br $>
（格式化！）
-
Richard Heathfield
" ; Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上述域名中， - www。
printf can print void pointer values if you use %p, but that''s it. So the
above line should read:
printf(" %p -- %p -- %p\n", (void *)(*p)[0],
(void *)(*p)[1],
(void *)(*p)[2]);
(Format to taste!)
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
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