What is the meaning of **(p2[1][0] 1)-CodePudding

I have just started diving into pointers in C and I can't understand why 14 is the output.

I have the given code:

#include<stdio.h>
#include<stdlib.h>
#define N 3
typedef int(*MyType1)[N];
typedef int(*MyType2)[N][N];

int main(int argc, int *argv[]){

    int m1[][N] = { 21,122,-13,14,56,36,17,78,92 };
    int m2[][N] = { 11,9,43,17,32,99,127,34,69 };
    MyType1 p1[] = { m1   1,m2   2,m1 2,m2,m1,m2 1};
    MyType2 p2[] = { &m2, &m1 };

    printf("%d\n", **(p2[1][0]   1));

    return 0;
}

My thought was: **(p2[1][0] 1) is equivalent to **(&m1[0] 1).

Then, &m1[0] is the address of the first element in m1 (21), plus one increments the address to the number next to it (122).

I noticed that every plus one accessed the n 3 number in m1 (goes down a row in m1?) --> The question is why.

CodePudding user response：

Well...

p[1] is a pointer to m1.

m1 might also have been initialized: {{21,122,-13},{14,56,36},{17,78,92}}.

So p[1][0] is a pointer to the first array: {21, 122, -13} but then we do some pointer arithmetic and point to the next array: {14,56,36} and when we dereference that we get the first element which is... 14.