In C, why can't a const variable be used as an array size initializer? [duplicate]

Question

In the following code, const int cannot be used as an array size:

const int sz = 0;
typedef struct
{
   char s[sz];
} st;

int main()
{
   st obj;
   strcpy(obj.s, "hello world");
   printf("%s", obj.s);
   return 0;
}

Answer 1

In C, a const-qualified variable is not a constant expression¹. A constant expression is something that can be evaluated at compile time - a numeric literal like 10 or 3.14159, a string literal like "Hello", a sizeof expression, or some expression made up of the same like 10 + sizeof "Hello".

For array declarations at file scope (outside the body of any function) or as members of struct or union types, the array dimension must be a constant expression.

For auto arrays (arrays declared within the body of a function that are not static), you can use a variable or expression whose value isn't known until runtime, but only in C99 or later.

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1. C++ is different in this regard - in that language, a const-qualified variable does count as a constant expression.