我这两个程序看起来一样，但其中一个导致错误

Question

here are my two programs and have a small difference. the first program compiles without error but the second program is giving error [enter image description here][1]

Program 1:

#include<stdio.h>
int main() {
    int a[][4] = { 5,7,5,9,4,6,3,1,2,9,0,6 };

    int *p;
    int(*q)[4];
    p = (int*)a;
    q = a;
    printf("%u %u\n", p, q);
    p++;
    q++;
    printf("%u %u\n", p, q);
    return 0;
}

this program compiles without any errors

Program 2:

#include<stdio.h>
int main(){
    int a[][4]={5,7,5,9,4,6,3,1,2,9,0,6};

    int *p;
    int *q[4];
    p=(int*)a;
    q=a;

    printf("%u %u\n",p,q);

    p++;
    q++;

    printf("%u %u\n",p,q);
    return 0;
}

program 2 shows error in the line 8 and 13 WHY?

Answer 1

The difference in the programs are int (*q)[4]; versus int *q[4];

So let's see what https://cdecl.org/ says about the types.

First code example

int (*q)[4];

declare q as pointer to array 4 of int

So here q is a pointer and therefore you can assign to it - like q=a;

Second code example

int *q[4];

declare q as array 4 of pointer to int

So here q is an array and therefore you can not assign to it, ie q=a; is illegal.

For the second code example gcc gives the error:

error: assignment to expression with array type
     q=a;
      ^

which actually says the same, ie that you are assigning to something being an array (and that is an error, ie illegal).

Answer 2

It is because arrays are not assignable. q in the second program is defined as an array of pointers, so when you do: q = a; , this is not valid in C. Likewise, when you write q++ , this is equivalent to q = q + 1; , so same kind of error.