Question

#include<stdio.h>
void main() {
 int s[4][2]={
               {1,2},
               {3,4},
               {5,6},
               {7,8}
             };
int (*p)[2]; // what does this statement mean? (A)
int i,j,*pint;

for(i=0;i<=3;i++) {
 p=&s[i];
 pint=(int*)p;  // what does this statement mean? (B)
 printf("\n");
  for(j=0;j<=1;j++) {
    printf("%d",*(pint+j));
  }
}

I cannot understand Statement ‘A’ and ‘B’ .How and what has been done?
please explain this very clearly.

Answer 1

Statement A is a declaration

int (*p)[2];
      ^      p is
int (*p)[2];
     ^       p is a pointer
int (*p)[2];
        ^    p is a pointer to an array
int (*p)[2];
         ^   p is a pointer to an array of 2
int (*p)[2];
^^^          p is a pointer to an array of 2 int

Statement B is an assignment expression with an embedded cast

pint=(int*)p;
           ^  take the value in p (of type "pointer to array of 2 ints")
pint=(int*)p;
     ^^^^^^   take the value in p, convert it to 'pointer to int'
              even if it doesn't make sense to do so
pint=(int*)p;
^^^^^         take the value in p, convert it to 'pointer to int'
              and put the resulting value (whatever that may be) in pint

Casts are bad. Avoid casts as much as possible.
^{(*) except in very particular circumstances like <ctype.h> or variadic functions or …}