ANOTHER CASE OF MANIPULATING AN ARRAY USING POINTERS

Introduction

You can put values in the memory locations by using pointers, but you cannot assign the memory location to an array to access those values because an array is a pointer constant.

Program

#include  void printarr(int a[]); void printdetail(int a[]); void print_usingptr_a(int a[]); main() {     int a[5];     int *b;     int *c;     for(int i = 0;i<5;i++)     {         a[i]=i;     }     printarr(a);     *b=2;            \\ A     b++;             \\ B     *b=4;            \\ C     b++;     *b=6;            \\ D     b++;     *b=8;            \\ E     b++;     *b=10;     b++;     *b=12;     b++;     a=c; //error     \\F     printarr(a);   } void printarr(int a[]) {     for(int i = 0;i<5;i++)     {         printf("value in array %d\n",a[i]);     } } void printdetail(int a[]) {     for(int i = 0;i<5;i++)     {         printf("value in array %d and address is %16lu\n",a[i],&a[i]);     } }  void print_usingptr_a(int a[]) {      for(int i = 0;i<5;i++)     {         printf("value in array %d and address is %16lu\n",*a,a); \\ F         a++; // increase by 2 bytes         \\ G     } } 

Explanation

1. You can assign a value at the location specified by b using statement A.

2. Using statement B, you can point to the next location so that you can specify a value at that location using statement C. Using this procedure, you can initialize 5 locations.

3. You cannot assign the starting memory location as given by statement F to access those elements because a is a pointer constant and you cannot change its value.

4. The function print_usingptr_a works correctly even though you are writing a++. This is because when you pass a as a pointer in an actual parameter, only the value of a is passed and this value is copied to the local variable. So changing the value in the local variable will not have any effect on the outside function.

Point to Remember

The array limit is a pointer constant and you cannot change its value in the program.