Variable Length Array (VLA) in C++ compilers

问题:

As we already know, VLA (standardized in C99) are not part of the standard in C++.

So the code below is "illegal" in C++:

void foo(int n) {
  int vla[n];
  for (int i = 0; i < n; ++i) {
    vla[i] = i;
  }
}

Despite of that the compiler (g++ and clang++) accepts the code as valid syntax, producing just a warning in case -pedantic flag is enable.

ISO C++ forbids variable length array ‘vla’ [-Wvla]

My questions are:

• Why does the compiler accept that declaration?
  The compiler cannot just reject an array in which length [is-no-know-at-compile-time]?
  Is there a sort of compatibility syntax rule to follow?

• What does the standard say about?
  From the assembly code produced I see the compiler writes in the stack in the loop, like a normal array, but I cannot find anything about the standard behaviour.

回答1:

Why does the compiler accept that declaration?

Because its authors chose to make it do so.

GCC in particular allows, by default, a lot of non-standard stuff that was historically accepted by old C compilers. They like "compatibility" in that sense.

What does the standard say about [it]?

Precisely what the warning states it says about it: ISO C++ forbids variable length arrays.

C++ does not have VLAs.

Where you see one being accepted, it is a compiler extension; to find out how that compiler implements such an extension, you would have to ask the compiler's authors (or examine its source, if applicable).

回答2:

The standard requires that a conforming compiler must "issue a diagnostic" when it encounters something that is illegal. Having done that, it's free to continue to compile the code with an implementation-specific meaning. (Note that "with an implementation-specific meaning" is a polite form of "with undefined behavior").