L Value required as increment operator - C

Question

I just got a question from my Friend.

#include<stdio.h>

void fun(int[][3]);

int main(void){
    int a[3][3]={1,2,3,4,5,6,7,8,9};

    fun(a);
    printf("\n%u",a);
    a++;//Ques 1

    printf("\n%u",a);
    printf("%d",a[2][1]-a[1][2]);

    return 0;
}

void fun(int a[][3]){
    ++a;//Ques 2
    a[1][1]++;
}

The line Ques 1 will throw an error of L value as 'a' is the name of a two dimensional array. But this is not happening with the case of line Ques 2.

Can any one clear this doubt?

Answer 1

In Ques 1, a is an array and it will be converted to a non-lvalue pointer when used as operand of ++ operator and emit compile error.

In Ques 2, the argument int a[][3] is equivalent to int (*a)[3] and ++a is an increment of pointer variable, which is acceptable.

Quote from N1570 6.7.6.3 Function declarators (including prototypes), pargraph 7:

A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation.