问题
In this excellent answer by AndreyT, he explains that in C, when a function needs an array whose dimension is known at compile-time, it's a major technique-level error to declare
void process_array(int *ptr, size_t plen);
instead of
void process_array(int (*arr_ptr)[10]);
Furthermore, he opines that many programmers are oblivious to the second option and know only about the first. One of the reasons, he writes, for this behaviour is when an array needs to be dynamically allocated and passed to the second version, programmers don't know how to do it; they're so accustomed to int *p = malloc(sizeof(*p) * 10) which returns an int *. In C, the way to do it is, as he shows
int (*arr_ptr) [10] = malloc(sizeof(*arr_ptr));
This got me thinking on how one would do the same in C++. I know we've std::array, std::vector, etc. but I'm interested in understanding new's usage. So I tried doing this:
typedef int Array10[10];
Array10 *p = new Array10; // error: cannot convert ‘int*’ to ‘int (*)[10]’
When I change p's type to int* the compiler (GCC 4.8.1) is happy. I looked up C++11 standard (draft n3337, §5.3.4/5) to understand this further, it says:
When the allocated object is an array (that is, the noptr-new-declarator syntax is used or the new-type-id or type-id denotes an array type), the new-expression yields a pointer to the initial element (if any) of the array. [Note: both new int and new int[10] have type int* and the type of new int[i][10] is int (*)[10] — end note]
I understand that new int [10] is in action; what I get back should be freed using delete [] p and not delete p. However what I need seems to be the latter, where the allocation is not an array of integers but an array itself, as a single/whole object.
Is there a way to do it? Or my trying to do this in itself shows a misunderstanding of the C++ type system? Or it is right but is simply not allowed by the standard?
Aside: When a function takes an array whose size is fixed, as in option 2, IMHO the best thing to do for the calling function is to declare an automatic array instead of resorting to dynamically allocating one and worrying about clean-up.
回答1:
Yes, in fact it is suggested in the standard text you quoted:
int (*arr_ptr)[10] = new int[1][10];
The pointer returned by the new-expression has type int (*)[10], i.e. pointer to array of 10 ints. You need delete[] to delete it.
回答2:
If I have understood correctly what you want then you can use std::array. For example
{
typedef std::array<int, 10> Array10;
Array10 *p = new Array10;
delete p;
}
来源：https://stackoverflow.com/questions/19467909/will-the-new-expression-ever-return-a-pointer-to-an-array