C++ Arrays & pointers -

can me following inequalities arrays:

#include <stdio.h> #define prd(a) printf("%d", (a)) #define nl printf("\n")  int a[] = {0,1,2,3,4};  int main () {   int i;   int*p;    (i=0; i<=4; i++)     prd(a[i]);   nl;    (p=&a[0]; p<=&a[4]; p++)     prd(*p);   nl;    (p=&a[0]; i=0; p+i<=a+4; i++)     prd(p[i]);   nl;    (p=a, i=0; p+i<=a+4; p++, i++)     prd(*(p+i));   nl; 

basically dont understand 4 loops, please kindly explain me!

a detail here, "inequalities" you're asking about pointers, not arrays. think of comparing memory locations, , following easier follow.

• first 1 should pretty straightforward: print each element of array in order (a[0], a[1], etc.)
• in number 2, p address of element 0 of a. p++ increases address address of next element in array.the termination condition can read "p less or equal address of element number 4 of array". put values , make concrete: imagine &a[0] 200, consequence &a[4] 216, p incremented becomes 204, 208, 212, 216, , 220. when becomes 220 condition becomes false, , loop ends.
• number 3's left side similar instead of increasing p adds i p. note magic of pointer arithmetic p + 1 p + i*(sizeof(*p)). sequence same in case 2. btw, initialization should p = &a[0], i=0; (note comma before i). right side (a+4), plain means address of first element of (this c equivalency between pointers , arrays, a topic full of subtleties , pain). a + 4 equivalent &a[4].
• number 4 there mess head. information above should enough translate initialization , condition; having both i++ , p++ means advancing 2 elements each iteration.

the use of preprocessor here obscuring gets called, try expanding manually follow more easily. , prd should adding space after each number easier reading.