谢谢你对我int(*p)[5]这个问题的指教，在几个小时前，热情依然已经告诉我怎么回事了，我先前只是知道它的字面意思，现在知道它怎么用了，呵呵。

不管怎么说，谢谢你了。

不过其他的问题仍然是不存在的。

我只是在不清楚int(*p)[5]是个什么东西的情况下，才以为要动态分配内存，因为调试时它老是0x000000，
现在误会搞清楚了，但是和我们先前的观点是没有关系的，我对int(*p)[5]的误解，并不表示其他指针知识的缺乏，今天感谢热情依然和穆兄你，我知道了int(*p)[5]的完全用法。

你认为int *pt = new int[5];那个例子pt==0x000000有问题？
非要void*p=new int[5];才没问题？那么我想问下：(new int[5])这个东西是什么类型？
要记得，其实new int[5];相当于是返回一个int指针，只是因为void指针可以指向任何类型的对象，才能void*p=new int[5];如果有争议：请首先回答(new int[5])这个东西是什么类型？

I want say int (*)p[5] != int p[5]; they are not same.

first one ist point array, that mean an array for pointer, here will a place for 5*4 Byte allocated, notice this 20 Byte is not dynamicaly allocation. and then in p[0] ... p[4] you can assign a pointer there, that mean, p[0] can save a pointer value, p[1] too,... p[4] too.

And now look at int p[5], this is an integer array, that mean, in this array can only integer value saved.

小小k又开始卖弄他的英文了，呵呵
有设么疑问，编译成汇编语言不都全部解决了么？

哈哈,我还想听一下kai兄说一说关于Reference counting

正想学汇编

我会，以前很多C语言不明白的地方，用WIN－TC编译成汇编一看就全清楚了

就是因为这样我才想学汇编

you have all made a mistake. That is you all think the return type from new is void. That is wrong.

take a look there, that is a link from msdn: http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vclang/html/_pluslang_new_operator.asp

The correct is, the return type from new is a suitable type. And the return type from malloc is really (void *) . So please pay attention to this difference.

That is why int * p = new int; you don't need write so int * p = (int *) new int; of course you can write your code like this, but unnecessary. But when you use malloc then you must use type cast, why? Because the return type from malloc is (void *), so we must use type cast, code like this: int * p = (int *) malloc( sizeof (int));

[此贴子已经被作者于2006-7-24 22:13:30编辑过]