Variable sized array on the stack [duplicate]

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g++ variable size array no warning? 1 answer

It is my understanding that in C and C++ , we create data-structures whose size is known at compile time on the stack and we use the heap (malloc-free / new-delete) for things whose size is not known at compile time and is decided at run-time.Why then , does my g++ compiler allow me to do something like the following code snippet.

int main(void)
{
    int n ;
    cin >> n ; // get the size of array.
    int arr[n] ; // create a variable sized array.
    .... do other stuff ...
}

Specifically , in the case of an array:
An array is assigned a contiguous block of memory on the stack and there are variables above and below it on the stack , so the array's size must be known so that the variable above the array on the stack , the array itself , and the variables below the array on the stack can all fit into memory neatly. How then are variable sized arrays on the stack implemented ? Why are they even necessary? Why can't we just use the heap for variable sized buffers?

EDIT:
I understand from the comments , that C and C++ have different rules regarding whether VLA's are standard or not and also Neil Butterworth's comment that it is generally not a good idea to ask a question about two languages at once. Thank you people , so I am removing the C tag from my question , as I intended to mainly ask about C++ , as evident by the code snippet syntax. Sorry for the confusion , and thanks for the responses.

They aren't allowed in standard C++, but they are allowed in standard C, and g++ allows them in C++ as well, as a language extension.

How then are variable sized arrays on the stack implemented?

See this question. The gist is that the size (n) gets evaluated and stored in a compiler-generated local variable, and then that much stack space is allocated. There's no law of physics that says the size of a stack variable has to be known at compile-time - it's merely simpler that way.