CODEWITHSUNDEEP
coptimization
Prasoon Saurav
Prasoon Saurav

Reputation: 92864

Pointer to array and related expression evaluation

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.

Upvotes: 2

Views: 159

Answers (1)

markgz
markgz

Reputation: 6266

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

  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.

Upvotes: 2

Related Questions