C Exercises: Show a pointer to an array which contents are pointer to structure

C Pointer : Exercise-20 with Solution

Write a program in C to show a pointer to an array which contents are pointer to structure.

Pictorial Presentation:

Sample Solution:

C Code:

#include <stdio.h>
struct employee
{
char *empname;
int empid;
};

int main()
{
 	printf("\n\n Pointer : Show a pointer to an array which contents are pointer to structure :\n"); 
	printf("-----------------------------------------------------------------------------------\n");

	static struct employee emp1={"Jhon",1001},emp2={"Alex",1002},emp3={"Taylor",1003};
	struct employee(*arr[])={&emp1,&emp2,&emp3};
	struct employee(*(*pt)[3])=&arr;

	printf(" Exmployee Name : %s \n",(**(*pt+1)).empname);
	printf("---------------- Explanation --------------------\n");
	printf("(**(*pt+1)).empname\n");
	printf("= (**(*&arr+1)).empname   as pt=&arr\n");
	printf("= (**(arr+1)).empname     from rule *&pt = pt\n");
	printf("= (*arr[1]).empname       from rule *(pt+i) = pt[i]\n");
	printf("= (*&emp2).empname        as arr[1] = &emp2\n");
	printf("= emp2.empname = Alex       from rule *&pt = pt\n\n");
	printf(" Employee ID :  %d\n",(*(*pt+1))->empid);	
	printf("---------------- Explanation --------------------\n");
	printf("(*(*pt+1))-> empid\n");
	printf("= (**(*pt+1)).empid     from rule -> = (*).\n");
	printf("= emp2.empid = 1002\n");
	printf("\n\n");
	return 0;
}

Sample Output:

 Pointer : Show a pointer to an array which contents are pointer to structure :                               
-----------------------------------------------------------------------------------                           
 Exmployee Name : Alex                                                                                        
---------------- Explanation --------------------                                                             
(**(*pt+1)).empname                                                                                           
= (**(*&arr+1)).empname   as pt=&arr                                                                          
= (**(arr+1)).empname     from rule *&pt = pt                                                                 
= (*arr[1]).empname       from rule *(pt+i) = pt[i]                                                           
= (*&emp2).empname        as arr[1] = &emp2                                                                   
= emp2.empname = Alex       from rule *&pt = pt                                                                 
                                                                                                              
 Employee ID :  1002                                                                                          
---------------- Explanation --------------------                                                             
(*(*pt+1))-> empid                                                                                            
= (**(*pt+1)).empid     from rule -> = (*).                                                                   
= emp2.empid = 1002 

Flowchart:

C Programming Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous:> Write a program in C to show a pointer to union.
Next: Write a program in C to print all the alphabets using pointer.