Program Ape--c Language Details 13 (two-dimensional array and pointers, &*a[i][0] understanding, array 1[e] and e[1] you probably haven't seen it.

Main content: two-dimensional arrays and pointers, &*a[i][0] understanding, arrays 1[e] and e[1]

#include <stdio.h> #define NUM_ROWS 10#define num_cols 10int Main (int argc, char **argv) {    int a[nu M_rows][num_cols], *p, i = 0; A is interpreted as a pointer to an integer pointer that is int * *     int C, d=2,*test, e[2] = {4,5},f[2][2] = {{11,22},{33,44}}; e understood as int *     test = &d;//    c = *&test; As with & (*test) results, * and & same priority, right-to-left     c = &*test; Attention and ++*test difference, here is a bit not understand, why & (*test) not error, I                 //sense is *test equivalent to D, not a specific value, and then & (*test) equals & (d)                 //So it is confirmed that the following pointer operation & and * can be canceled//    c = ++*test;     printf ("1[e] =%d\n", 1[e]); 1[e] and e[1], as for the compiler e[1] equivalent to * (E + 1) ==>* (1 + E)            & Nbsp;                   //is equivalent to 1[e], but good programming habits do not write this     printf ("c =%d\n", c);    printf (" D is the address =%d\n ", &d);//    test = &*e[0];  //This error (*&e[0] compile through but run error), the pointer can be & and * Cancel, e[0] is constant,                     //below f[1] indicates the second row//    printf ("constant d =%d\n", *test);    test = & *f[1];    printf ("D =%d\n", *test);    for (i = 0; i < num_rows; i++)     {          /* causes p to point to a row in a two-dimensional array with P = &a[i][0] equivalent to P = a[i]          * Two-dimensional arrays in C language storage by row          * The magic formula between array subscript and pointer arithmetic operations          * for any array a , A[i] equivalent to * (a+1) .         * because & and * can be canceled, then &a[i][0] equivalent to & (* (a[i]+0)) ==> equivalent to &*a[i] ==>a[i]         */&Nbsp;       for (p = a[i]; p < a[i] + num_cols; p++)         {&nbsp ;           *p = i;             printf ("%3d", *p);        }        printf ("\ n");             }     return 0;}

Output:

Program Ape--c Language Details 13 (two-dimensional array and pointers, &*a[i][0] understanding, array 1[e] and e[1] you probably haven't seen it.