What will be the output of the following program? #include   void main ()   {       int a[10] = {100, 206, 300, 409, 509, 601}; //Line 1       int *p[] = {a, a+1, a+2, a+3, a+4, a+5}; //Line 2       int **pp = p; //Line 3       pp++; // Line 4       printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 5       *pp++; // Line 6       printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 7       ++*pp; // Line 8       printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 9       ++**pp; // Line 10        printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 11   } I don't need screenshot I need explanation of the output.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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What will be the output of the following program? #include<stdio.h>  

void main ()  

{  

    int a[10] = {100, 206, 300, 409, 509, 601}; //Line 1  

    int *p[] = {a, a+1, a+2, a+3, a+4, a+5}; //Line 2  

    int **pp = p; //Line 3  

    pp++; // Line 4  

    printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 5  

    *pp++; // Line 6  

    printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 7  

    ++*pp; // Line 8  

    printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 9  

    ++**pp; // Line 10   

    printf("%d %d %d\n",pp-p,*pp - a,**pp); // Line 11  

}

I don't need screenshot I need explanation of the output.

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