c - How pointer to array works? -

int s[4][2] = {                   {1234, 56},                   {1212, 33},                   {1434, 80},                   {1312, 78}               };  int (*p)[1]; p = s[0];  printf("%d\n", *(*(p + 0))); // 1234 printf("%d\n", *(s[0] + 0)); // 1234 printf("%u\n", p);           // 1256433(address of s[0][0]) printf("%u\n", *p);          // 1256433(address of s[0][0]) 

can explain why doing *(*(p + 0)) prints 1234, , doing *(s[0] + 0) prints 1234, when p = s[0] , why p , *p gives same result?

thanking in anticipation.

this way arrays work in c -- arrays not first class types, in can't them other declaring them , getting size. in other context, when use expression type array (of anything) silently converted pointer array's first element. referred array "decaying" pointer.

so lets @ statements 1 one:

p = s[0]; 

here, s has array type (it's int[4][2] -- 2d int array), silently converted pointer first element (an int (*)[2], pointing @ word containing 1234). index [0] adds 0 * sizeof(int [2]) bytes pointer, , dereferences it, giving int [2] (1d array of 2 ints). since array, silently converted pointer first element (an int * pointing @ 1234). note same pointer before index, pointed @ type different.

you assign int * p, declared int (*)[1]. since c allows assigning pointer other pointer (even if pointed @ types different), works, reasonable compiler give type mismatch warning.

p points @ word containing 1234 (the same place pointer s points at)

printf("%d\n", *(*(p+0))); 

this first adds 0*sizeof(int[1]) p , dereferences it, giving array (int[1]) decays pointer first element (an int * still pointing @ same place). pointer dereferenced, giving int value 1234 printed.

printf("%d\n", *(s[0]+0)); 

we have s[0] again via multiple decay , dereference process noted in description of first line, becomes int * pointing @ 1234. add 0*sizeof(int) it, , dereference, giving integer 1234.

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

p pointer, address of pointer printed.

printf("%u\n",*p) 

p dereferenced, giving int [1] (1d integer array) decays pointer first element. pointer printed.