Why is new int[n] valid when int array[n] is not?

Question

For the following code:

foo(int n){
    int array[n];
}

I understand that this is invalid syntax and that it is invalid because the c++ sta

Answer 1

This is because the C++ language does not have the C feature introduced in C99 known as "variable length arrays" (VLA).

C++ is lagging in adopting this C feature because the std::vector type from its library fulfills most of the requirements.

Furthermore, the 2011 standard of C backpedaled and made VLA's an optional feature.

VLA's, in a nutshell, allow you to use a run-time value to determine the size of a local array that is allocated in automatic storage:

int func(int variable)
{
   long array[variable]; // VLA feature

   // loop over array

   for (size_t i = 0; i < sizeof array / sizeof array[0]; i++) {
     // the above sizeof is also a VLA feature: it yields the dynamic
     // size of the array, and so is not a compile-time constant,
     // unlike other uses of sizeof!
   } 
}

VLA's existed in the GNU C dialect long before C99. In dialects of C without VLA's, array dimensions in a declaration must be constant expressions.

Even in dialects of C with VLA's, only certain arrays can be VLA's. For instance static arrays cannot be, and neither can dynamic arrays (for instance arrays inside a structure, even if instances of that structure are allocated dynamically).

In any case, since you're coding in C++, this is moot!

Note that storage allocated with operator new are not the VLA feature. This is a special C++ syntax for dynamic allocation, which returns a pointer type, as you know:

int *p = new int[variable];

Unlike a VLA's, this object will persist until it is explicitly destroyed with delete [], and can be returned from the surrounding scope.