Incomplete array type int (*p_arr)[ ]?

Question

When I compiled the following code with gcc -Wall -pedantic -ansi -std=c89, it compiled successfully without giving an error at the pointer assignment. Note that I convert from int (*)[4] to int (*)[].

int arr[4];
int (*p_arr)[] = &arr;

Assuming that there is some reason for allowing this (incompatible?) assignment, when I try to use it, the compiler gives incomplete type error error: invalid application of ‘sizeof’ to incomplete type ‘int[]’.

(void) sizeof(*p_arr);

This error makes me think what is the use of allowing the previous pointer assignment p_arr = &arr? Is this assignment allowed as per the standard?

I have used incomplete struct/union type (usually for forward declaration) and also come across the error incomplete array element type. But this incomplete array type is new to me. Is it possible in C standard and has a use case?

Answer 1

This assignment didn't give any error because their types are compatible because arrays of unknown bound are compatible with any array of compatible element type. (For reference)-

int (*p_arr)[] = &arr;

But gives error on passing it as operand to sizeof operator because *p_arr is of incomplete type and you are not supposed to use incomplete types as operands to sizeof operator.

N1570 6.5.3.4

1 The sizeof operator shall not be applied to an expression that has function type or an incomplete type, to the parenthesized name of such a type, or to an expression that designates a bit-field member[...].

Now what you can use it, here is a simple example -

#include <stdio.h>

int main(void){
    int arr[4]={1,2,3,4};
    int a[6]={1,2,3,3,1,1}; 
    int (*p_arr)[] = &arr;
    for(int i=0;i<4;i++)
       printf("%d",(*p_arr)[i]);
    printf("\n");
    p_arr=&a;
    for(int i=0;i<6;i++)
       printf("%d",(*p_arr)[i]);
    return 0;
}