有趣的指针问题？ [英] interesting pointer problem?
问题描述
我发现了一个有趣的指针问题。
指针是什么意思？
EXTERN short（* blocks）[64];
malloc是什么意思？
blocks =
（短（*）[64]）malloc（count * sizeof（short [ 64]））;
如有任何建议，将不胜感激！
祝你好运，
Davy
I found a interesting pointer problem.
What does the pointer mean?
EXTERN short (*blocks)[64];
And what does the malloc mean?
blocks =
(short (*)[64])malloc(count*sizeof(short [64]));
Any suggestions will be appreciated!
Best regards,
Davy
推荐答案
2005年9月6日星期二01:40:01 -0700，Davy写道：
On Tue, 06 Sep 2005 01:40:01 -0700, Davy wrote:
我找到了一个有趣的指针问题。
指针是什么意思？
EXTERN short（* blocks）[64];
EXTERN不是标准标识符，它可能被定义为一个宏
某处扩展到extern或什么都没有。
blocks是指向64个短裤数组的指针。
malloc是什么意思？
blocks =
（短（*）[64]）malloc （count * sizeof（short [64]））;
I found a interesting pointer problem.
What does the pointer mean?
EXTERN short (*blocks)[64];
EXTERN isn''t a standard identifier, it is probably defined as a macro
somewhere expanding to extern or nothing.
blocks is a pointer to an array of 64 shorts.
And what does the malloc mean?
blocks =
(short (*)[64])malloc(count*sizeof(short [64]));
这实际上分配了一个2D数组（实际上是数组数组）的数量
* 64个短路，其元素可以使用块[a] [b]访问。假设
当然，malloc（）调用成功。
注意这是C和C ++有不同方法的情况。在C中你/ b
将倾向于在没有演员的情况下使用malloc（）调用，在C ++中你会使用新的运算符来使用
。
劳伦斯
This allocates in effect a 2D array (in fact an array of arrays) of count
* 64 shorts whose elements can be accessed using blocks[a][b]. Assuming
the malloc() call succeeds of course.
Note this is a case where C and C++ have different approaches. In C you
would tend to use the malloc() call without the cast, in C++ you would
tend to use the new operator.
Lawrence
Davy写道：
Davy wrote:
我发现了一个有趣的指针问题。
指针是什么意思？
EXTERN short（* blocks）[64];
假设EXTERN只是#defined与extern相同，这意味着声明了一个名为''blocks''的
变量（未定义）这是指向64个短值的
数组的指针。
malloc是什么意思？
blocks =
（短（*） [64]）malloc（count * sizeof（short [64]））;
I found a interesting pointer problem.
What does the pointer mean?
EXTERN short (*blocks)[64];
Assuming EXTERN is just #defined to be the same as extern, it means that a
variable named ''blocks'' is declared (not defined) that is a pointer to an
array of 64 short values.
And what does the malloc mean?
blocks =
(short (*)[64])malloc(count*sizeof(short [64]));
它分配''count''64个数组的短数并分配地址第一个
一个到''块''。
It allocates ''count'' arrays of 64 short and assigns the address of the first
one to ''blocks''.
谢谢你的帮助！
所以阻止是一个指向动态二维数组的指针，是吗？
我很困惑
"短（*块）[64];"
和短[64]（*块）;"
和short * blocks [64];" ;.
它们的区别是什么？
祝你好运，
Davy
Hi,
Thank you for your help!
So the "blocks" is a pointer to a dynamic 2-D array, is it?
I am confused with
" short (*blocks)[64];"
and " short [64] (*blocks);"
and "short *blocks [64];".
What''s their difference?
Best regards,
Davy
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