为什么不能将一个数组分配给另一个数组？

Question

I can't seem to wrap my head around the reason the following code won't compile.

In my header file I declared an array as a static class member:

class foo {
private:
#define SIZE 50
static char array[SIZE];
// further code goes here
}

In the implementation, I have to initialize the array.

char foo::array[SIZE] = new[] char[SIZE];

This yields me an error everytime - the compiler says:

cannot convert from 'char *' to 'char [50]'

Why does the compiler interpret new[] char[SIZE] as char* ?

Answer 1

1. There's no such thing as new [] . You need to remove the [] .
2. Because ::operator new returns a pointer, not an array. Pointers are not arrays, and trying to treat the two interchangeably will result in pain. Just because arrays will decay into pointers doesn't mean that arrays are pointers.
3. foo::array[SIZE] is already static -- there's no need to allocate storage for it in any case.
4. This looks like you're coming from a Java or C# background, where arrays are reference types. Arrays are not reference types in C++. When you declare the array the storage for it is implicit. In your case the storage is going to be where all the other static s are; if the array was just written that way without static in a function then it would be allocated on the stack.

Answer 2

In C++ array is almost like constant pointer ( char * const , not char const * ). If you assign pointer to another pointer you just copy the address, not the value.

As Billy said in your case you don't need to allocate memory for the array, it's already allocated, because you declared it as static array with fixed length. You use operator new if you want to create object on heap or array with length known in runtime. You can to this like this:

class A{
public:
    A(int s): array_(new char[s]), length_(s) {}
    ~A() { delete [] array_; }
private:
    char * const array_;
    int length_;
};