一个指向数组的指针？ [英] a pointer to an array?
问题描述
大家好，全部：
最近我在comp.lang.c上找到了这条评论" Answers to FAQ" （6.13）：
如果你真的需要声明一个指向整个数组的指针，请使用
之类的int（* ap）[N];其中N是数组的大小。 （参见
也问题1.21。）如果数组的大小未知，N可以在
原则中省略，但结果类型，指向数组的指针
未知大小，没用。
但是，当我尝试编写一个示例程序时：
int main（）
{
int（* arrptr）[5];
int arr [5] = {1,2,3,4,5};
arrptr = arr;
printf（"％d"，arrptr [0]）;
}
为什么不能工作？此外，有没有人知道
阵列指针的实际用法？
祝你好运，
blue
Hi, all:
Recently I found this comment on comp.lang.c "Answers to FAQ" (6.13):
If you really need to declare a pointer to an entire array, use
something like "int (*ap)[N];" where N is the size of the array. (See
also question 1.21.) If the size of the array is unknown, N can in
principle be omitted, but the resulting type, "pointer to array of
unknown size," is useless.
However, when I try to write a sample program like:
int main()
{
int (*arrptr)[5];
int arr[5] = {1,2,3,4,5};
arrptr = arr;
printf( "%d", arrptr[0] );
}
Why could it not work? Besides, does anyone know the real usage of
array pointers?
Best regards,
blue
推荐答案
blue schreef：
blue schreef:
全部：
最近我在comp.lang.cAnswers to FAQ中找到了这条评论。 （6.13）：
如果你真的需要声明一个指向整个数组的指针，请使用类似int（* ap）[N];之类的东西。其中N是数组的大小。
在C ++中我会推荐它。不幸的是，它无效C.
（另见问题1.21。）如果数组的大小未知，N可以原则上省略，但是结果类型，指向未知大小的数组的指针，没用。
但是，当我尝试编写一个示例程序时：
int main（）
{
int（* arrptr）[ 5];
int arr [5] = {1,2,3,4,5};
arrptr = arr;
printf（"％d"，arrptr [0]） ;
}
为什么它不起作用？
因为它不是C.
int main（）
{
int * arrptr;
int arr [5] = {1,2,3,4,5};
arrptr = arr;
printf（"％d"，arrptr [0]）;
}
此外，有没有人知道数组指针的实际用法？
Hi, all:
Recently I found this comment on comp.lang.c "Answers to FAQ" (6.13):
If you really need to declare a pointer to an entire array, use
something like "int (*ap)[N];" where N is the size of the array.
In C++ i would recommend it. Unfortunately, it''s not valid C.
(See also question 1.21.) If the size of the array is unknown, N can in
principle be omitted, but the resulting type, "pointer to array of
unknown size," is useless.
However, when I try to write a sample program like:
int main()
{
int (*arrptr)[5];
int arr[5] = {1,2,3,4,5};
arrptr = arr;
printf( "%d", arrptr[0] );
}
Why could it not work?
Coz'' it''s not C.
int main()
{
int *arrptr;
int arr[5] = {1,2,3,4,5};
arrptr = arr;
printf( "%d", arrptr[0] );
}
Besides, does anyone know the real usage of array pointers?
它们有很多用途。例如，printf通常需要一个
数组指针，因为它的第一个参数是％d。在你的情况下。你可以通过arrptr = arr + i计算指向数组中任何元素的指针。
除了它们总是指向某个有效的块
内存（只要它们在范围内），数组指针与
任何其他指针没有区别。
问候，
Kleuske
There are many uses for them. printf, for instance, usually takes an
array pointer as it''s first parameter, "%d" in your case. You can
calculate a pointer to any element in the array by arrptr=arr + i.
Except for the fact that they always point to some valid block of
memory (as long as they''re in scope), array-pointers do not differ from
any other pointer.
regards,
Kleuske
blue写道：
全部：如果你真的需要声明一个指向整个数组的指针，请使用类似int（* ap）[N];之类的东西。其中N是数组的大小。 （参见
同样问题1.21。）如果数组的大小未知，则可以省略N，但结果类型，指向未知大小的数组的指针， ;没用。
但是，当我尝试写一个示例程序时：
#include< stdio.h>
这是_not_无偿。
int main（）
{
int（* arrptr）[5];
int arr [5] = {1,2,3,4 ，5};
arrptr = arr;
这是什么？如果你有
int * p，我;
然后你用
p =& i;
所以，你的编译器应该告诉你
arrptr = arr;
是错误的。使用上面的方法，
arrptr =& arr;
是对的。
printf（"％d"，arrptr [0]） ;
而不是p [0]，你也可以写* p。所以，基本上，你是
正在尝试使用％d打印出一个数组。我希望你能够意识到这是一个愚蠢的想法。
你可能想要的是
arrptr [0] [0]
或者，在这种情况下，
（* arrptr）[0]
注意：您也可以使用arrptr访问数组N的一个
数组5的int - 也许这有助于你更好地理解
[0] [0]......
另一件事：为了使这个程序可移植，将''\ n''
附加到你的上一个输出，即％d \ n或者单独的putchar（''\ n''）;
你忘了
返回0; }
为什么它不起作用？此外，有没有人知道
数组指针的实际用法？
Hi, all:
Recently I found this comment on comp.lang.c "Answers to FAQ" (6.13):
If you really need to declare a pointer to an entire array, use
something like "int (*ap)[N];" where N is the size of the array. (See
also question 1.21.) If the size of the array is unknown, N can in
principle be omitted, but the resulting type, "pointer to array of
unknown size," is useless.
However, when I try to write a sample program like:
#include <stdio.h>
This is _not_ gratuitous.
int main()
{
int (*arrptr)[5];
int arr[5] = {1,2,3,4,5};
arrptr = arr;
What is that? If you have
int *p, i;
then you use
p = &i;
So, your compiler should tell you that
arrptr = arr;
is wrong. Using the above,
arrptr = &arr;
is right.
printf( "%d", arrptr[0] );
Instead of p[0], you can also write *p. So, essentially, you
are trying to print out an array using %d. I hope you are
aware that this is a stupid idea.
What you probably want is
arrptr[0][0]
or, in this case,
(*arrptr)[0]
Notes: You could also use arrptr to access an array N of an
array 5 of int -- maybe this helps you better to understand
the "[0][0]"...
Another thing: To make this program portable, append ''\n''
to your last output, i.e. "%d\n" or a separate putchar(''\n'');
You forgot to
return 0; }
Why could it not work? Besides, does anyone know the real usage of
array pointers?
见上文。是的。
干杯
Michael
-
电子邮件：我的是/ at / gmx / dot / de address。
See above. Yes.
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
kl ***** @ xs4all.nl 写道：
blue schreef：
blue schreef:
全部：
最近我在comp.lang上发现了这条评论。 c常见问题解答 （6.13）：
如果你真的需要声明一个指向整个数组的指针，请使用类似int（* ap）[N];之类的东西。其中N是数组的大小。
Hi, all:
Recently I found this comment on comp.lang.c "Answers to FAQ" (6.13):
If you really need to declare a pointer to an entire array, use
something like "int (*ap)[N];" where N is the size of the array.
在C ++中我会推荐它。不幸的是，它不是有效的C.
In C++ i would recommend it. Unfortunately, it''s not valid C.
这是错误的。
在comp.lang.c中，我不正式了解C ++，但正确地使用了
，这个绝对是有效的C - 之所以
，整个事情都进入了_comp.lang.c_常见问题解答。
另请参阅我对该主题的其他回复。
干杯
Michael
-
电子邮件：我的是/ at / gmx / dot / de address。
This is wrong.
In comp.lang.c, I do not officially know about C++ but, used
correctly, "this" is definitely valid C -- the reason why
the whole thing went into the _comp.lang.c_ FAQ.
See also my other reply to this thread.
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.
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