Pointer to array and related expression evaluation

Question

An array type decays to a pointer type when it is passed to a function

That means in

int func(int x[*p])

*p should not be evaluated as the declaration is equivalent to int func(int *x)

Does the same hold for pointer to arrays?

Here is the code

int *p=0;
void func(int (*ptr)[*p]) //A
{
   // code
}

int main()
{
   int arr[5][5];
   func(arr);
}

Is evaluation of *p at //A guaranteed by the Standard?

I tried with and without optimization on g++4.6. With optimizations enabled I don't get segfault. On clang the code is not giving any segfault even without any optimizations.

Isn't declaring void func(int (*ptr)[*p]) exactly the same as void func(int *ptr[])? — Bart
– Bart, Commented Feb 27, 2012 at 19:52
I had no idea that void func(int x[RANDOM_STUFF]) was valid! — Oliver Charlesworth
– Oliver Charlesworth, Commented Feb 27, 2012 at 20:57

markgz · Accepted Answer · 2012-02-27 23:02:19Z

2

From the C99 Standard Section 6.7.5.2 Array declarators, paragraph 1:

In addition to optional type qualiﬁers and the keyword static, the [ and ] may delimit an expression or *. If they delimit an expression (which speciﬁes the size of an array), the expression shall have an integer type. If the expression is a constant expression, it shall have a value greater than zero. The element type shall not be an incomplete or function type. The optional type qualiﬁers and the keyword static shall appear only in a declaration of a function parameter with an array type, and then only in the outermost array type derivation.

The expression *p evaluates to 0 and does not satisfy the requirements of the above paragraph, so the behavior is undefined.

answered Feb 27, 2012 at 23:02

markgz

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