Does “int size = 10;” yield a constant expression?

问题

The following code compiles under gcc 4.8 and Clang 3.2:

int main()
{
  int size = 10;
  int arr[size];
}

8.3.4/1 of the C++ Standard says that the size of an array must be an integral constant expression, which size does not seem to be. Is this a bug in both compilers, or am I missing something?

The latest VC++ CTP rejects the code with this interesting message:

error C2466: cannot allocate an array of constant size 0

The interesting part is how it seems to think that size is zero. But at least it rejects the code. Shouldn\'t gcc and Clang do the same?

回答1:

This is variable length arrays or VLA which is a C99 feature but gcc and clang support it as an extension in C++ while Visual Studio does not. So Visual Studio is adhering to the standard in this case and is technically correct. Not to say that extensions are bad, the Linux kernel depends on many gcc extensions, so they can be useful in certain contexts.

If you add the -pedantic flag both gcc and clang will warn you about this, for example gcc says (see it live):

warning: ISO C++ forbids variable length array 'arr' [-Wvla]
  int arr[size];
              ^

Using the -pedantic-errors flag will make this an error. You can read more about extensions in these documents Language Standards Supported by GCC and clangs Language Compatibility section.

Update

The draft C++ standard covers what is a integral constant expression in section 5.19 Constant expressions paragraph 3 and says:

An integral constant expression is an expression of integral or unscoped enumeration type, implicitly converted to a prvalue, where the converted expression is a core constant expression. [...]

It is not intuitively obvious from reading this what all the possibilities are but Boost's Coding Guidelines for Integral Constant Expressions does a great job of that .

In this case since you are initializing size with a literal using const would suffice to make it an integral constant expression (see [expr.const]p2.9.1) and also bring the code back to being standard C++:

const int size = 10;

using constexpr would work too:

constexpr int size = 10;

It would probably help to read Difference between constexpr and const.

For reference the equivalent section to 8.3.4 paragraph 1 in the C99 draft standard would be section 6.7.5.2 Array declarators paragraph 4 which says (emphasis mine):

If the size is not present, the array type is an incomplete type. If the size is * instead of being an expression, the array type is a variable length array type of unspecified size, which can only be used in declarations with function prototype scope;¹²⁴⁾ such arrays are nonetheless complete types. If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type.

来源：https://stackoverflow.com/questions/21273829/does-int-size-10-yield-a-constant-expression