是否可以在C ++中动态创建恒定大小的数组？

Question

First of all, I want to reassure you all that I am asking this question out of curiosity. I mean, don't tell me that if I need this then my design has problems because I don't need this in real code. Hope I convinced you :) Now to the question:

For most types T we can write

T* p = new T;

now what if T is an array type?

int (*p)[3] =  new ???; //pointer to array of 3 = new ???

I tried this:

typedef int arr[3];
arr* p = new arr;

but this doesn't work.

Is there any valid syntax for this or it is impossible in C++. If it is impossible, then why? Thanks

Edit : i am guessing I wasn't clear enough. I want to be able to use it in this situation:

void f(int(&)[3]);
int (*p)[3] = new ???;
f(*p);

Answer 1

要从new获得指向数组的指针，必须动态分配一个二维数组：

int (*p)[3] = new int[1][3];

Answer 2

The reason you can't do it is that new int[3] already allocates exactly what you want, an object of type int[3] . It's just that what the new-expression returns , is a pointer to its first element. 5.3.4/1:

If the entity is a non-array object, the new-expression returns a pointer to the object created. If it is an array, the new-expression returns a pointer to the initial element of the array.

Returning a pointer to the first element is what allows the 3 to be unknown until runtime, so I suppose that by knowing it in advance, you've tripped over flexibility that you aren't using.

I guess the ways around this are to reinterpret_cast back to the pointer type you want (not necessarily portable), or to allocate a struct containing an int[3] (and use a pointer to its data member).

[Edit: er, yeah, or FredOverflow's idea, which has neither disadvantage, but requires use of delete[] instead of delete .]

I guess the moral is, if you write templates that naively allocate some unknown type T with new , then the template won't work when someone passes an array type as T . You'll be assigning it to the wrong pointer type, and if you fix that (perhaps with auto ), you'll be deleting it wrongly.

Edit in answer to j_kubik's question:

Here's one way to distinguish between array and non-array types. If you write a function like this, that returns an object that holds the pointer and is capable of correctly deleting it, then you have a generic new/delete for any type T.

#include <iostream>

template <typename T>
void make_thing_helper(T *) {
    std::cout << "plain version\n";
}

template <typename T, int N>
void make_thing_helper(T (*)[N]) {
    std::cout << "array version\n";
}

template <typename T>
void make_thing() {
    make_thing_helper((T*)0);
}

int main() {
    typedef int T1;
    typedef int T2[3];
    make_thing<T1>();
    make_thing<T2>();
}

Answer 3

You could always use boost::array, which will be in C++0x. Otherwise, any solution will be awkward at best: arrays are broken in C, and C++ maintains compatilibity with C in this respect. Fred Overflow offered one solution; even easier (but syntactically noisy) would be to wrap the array in a struct: struct A { int arr[3]; }; and allocate and manipulate this.

Answer 4

You just do

int *p = new unsigned int [3]

You can then use *p as a pointer or an array ie *(p+1) or p[1]