Scope of variably sized array

Question

Is this always going to run as expected?

char *x;
if (...) {
    int len = dynamic_function();
    char x2[len];

    sprintf(x2, "hello %s", ...);

    x = x2;
}

printf("%s\n", x);
// prints hello

How does the compiler (GCC in my case) implement variably sized arrays, in each of C and C++?

Answer 1

No. x2 is local to the if statement's scope and you access it outside of it using a pointer. This results in undefined behaviour.

By the way, VLAs have been made optional in C11 and had never been part of C++. So it's better to avoid it.

Answer 2

The scope is explained here :

Jumping or breaking out of the scope of the array name deallocates the storage. Jumping into the scope is not allowed; you get an error message for it.

In your case the array is out of scope.

Answer 3

No, for two separate reasons:

C++: The code isn't valid C++. Arrays in C++ must have a compile-time constant size.

C: No, because the array only lives until the end of the block in which it was declared, and thus dereferencing x is undefined behaviour.

From C11, 6.2.4/2:

If an object is referred to outside of its lifetime, the behavior is undefined.

And 6.2.4/7 says that the variable-length array lives from its declaration until the end of its enclosing scope:

For such an object that does have a variable length array type, its lifetime extends from the declaration of the object until execution of the program leaves the scope of the declaration.