C Programming

Tree, Graphs are the types of nonlinear data structure.

i dnt have time to draw flow chart but i'll give the algorith..(only major steps)

1 :read n

2:m=n

3:if n=0 goto step 8,else goto step 4

4:r=n%10

5:rev=(rev*10)+r

6:n=n/10

7:goto step3

8:if m==rev,display "palindrome" else "not palindrome

9:stop

Here is a simple program that will tell you how to make an algorithm:

int main();

{

int length;

int width;

int total;

printf("What is the width: ");

scanf("%d", &width);

printf("What is the length: ");

scanf("%d", &length);

total = width * 2 + 2 * length; /*Here is the algorithm for finding the perimeter of a square*/

printf("The perimeter is: %d", total);

return 0;

}

Output:

What is the width: 32

What is the length: 55

The perimeter is: 174

Tomcat

#include<stdio.h>

int main(){

int num,r,sum,temp;

int min,max;

printf("Enter the minimum range: ");

scanf("%d",&min);

printf("Enter the maximum range: ");

scanf("%d",&max);

printf("Armstrong numbers in given range are: ");

for(num=min;num<=max;num++){

temp=num;

sum = 0;

while(temp!=0){

r=temp%10;

temp=temp/10;

sum=sum+(r*r*r);

}

if(sum==num)

printf("%d ",num);

}

return 0;

}

17

• ana

#include <iostream>

int main()

{

int i=0;

while( i<10 )

printf( "%d\n", ++i );

return( 0 );

}

The size or length of the list. For static, the size is a constant, while the size of a dynamic list may change over time.

The 7 weekdays is static (in size/length, though the content is static as well), while the questions and answers at answers.com are 2 dynamic lists (the sizes are not constants, although just growing)

0X at the beginning represent a number in the hexadecimal system of units.

FFFF is the hexadecimal equivalent of

i) 65535 in decimal system of units

ii) 1111111111111111 in binary system of units

Call By Value, the standard way of doing things in C and C++, is where a copy of an object is placed in the parameter stack. The called function can access and manipulate that copy at will, but it cannot change the original copy because it has no way of knowing where that original copy is located.

Call By Reference, on the other hand, is where the address of an object is placed in the parameter stack. Using extra syntax, the * or the ->, the called function can access and manipulate the original copy at will.

Try this for a very, very simple face....

String hair = " /////// ";

String eyes = " | o o | ";

String noseEars = "(| ^ |)";

String mouth = " | [ ] | ";

String chin = " ------ ";

System.out.println(hair); //prints out the hair

System.out.println(eyes); //prints out the eyes

System.out.println(noseEars); //prints out the nose and ears

System.out.println(mouth); //prints out the mouth

System.out.println(chin); //prints out the chin

A generalized linked list contains structures or elements with every one containing its own pointer. It's generalized if the list can have any deletions, insertions, and similar inserted effectively into it.

There are many limitations of arrays. The greatest advantage of an array of course is performance. Arrays are by far the fastest means of referencing collection of items.

That said, arrays suffer incredible limitations since they are generally very slow for adding items in a sorted manor. Unless you preallocate a large enough area of memory to store all possible items you wish to use, arrays have to be reallocated over and over. There are some hackish functions which can be used to exploit the MMU of the system to grow arrays by creating multiple logical contiguous regions of memory, but if this is performed on a fine grain, the logical descriptor tables of the system will grow and overall system performance will be negatively impacted.

Even though you appear to want to program plain-C, there is no reason you can't develop in an object oriented manor. There are many libraries available for C which provide high-performance variable length collections. I am not familiar with all of them, but as part of the GNOME toolkit, there is GLib (if I recall) which provides excellent implementations of many of them. Alternatively, if you're interested in learning how they work in order to correctly choose the right one for each task, you may want to get a copy of Donald Knuth's "The Art of Computer Programming, Volume 1" which is effectively a bible of sorts on this specific topic and is quite easy to read.

If you want to learn the ABCs and 123s without an overly prolonged investment in time, I recommend experimenting with an object oriented language with good class libraries like Java or my favorite, C#. They are both great starting points for experimenting with data structures.

Authors are far more concerned with the actual content than the layout and will use word processors to produce that content. Desktop publishing is more concerned with layout than with content.

import java.io.*;

public class Arms

{

public static void main(String args[]) throws Exception

{

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter the number to be printed: ");

int n=Integer.parseInt(br.readLine());

int num,sum,r,temp;

System.out.println("Armstrong numbers upto given limit are:\n");

for(num=1;num<n;num++)

{

temp=num;

sum =0;

while (temp!=0)

{

r=temp%10;

sum=sum+(r*r*r);

temp=temp/10;

}

if(sum==num)

{

System.out.println(num);

}

}

}

}

Regards

Ramakrishna Nallapati

Important. (Please, be more specific with your question, or else no-one could fathom what you wants to know.)

There needs to be a category for questions that as asked defy a sensible answer.

no, every function can not return a value. for example

void name()

{

cout<<"Hello world";

}

this function does not return any value due to the key word void that tells the compiler that the function does not returns a value.

i think a pass is reading the input file, i.e. the file in which the code is written in the source language.so in a single pass compiler, the input file is read only once and in doublepass compiler this is done 2 times

.

Because C is one of the high level programming languages. Other examples: Pascal, Fortran, List, JavaScript, APL

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#include<process.h>

#include<string.h>

#define MAX 20

typedef struct queue

{

int data;

struct queue *next;

}queue;

typedef struct stack

{

int data;

struct stack *next;

}stack;

typedef struct adj_list

{

int vertex;

struct adj_list *next;

}adj_list;

adj_list *G[MAX];

int queue_empty(queue *front,queue *rear)

{

if(front==NULL && rear==NULL)

return 1;

else

return 0;

}

queue* getnode()

{

queue *newnode;

newnode = (queue*)malloc(sizeof(queue));

newnode->next = NULL;

return newnode;

}

void enqueue(queue **front, queue **rear, int data)

{

queue *newnode;

newnode = getnode();

newnode->data = data;

if(queue_empty(*front,*rear))

{

*front = *rear = newnode;

}

else

{

(*rear)->next = newnode;

(*rear) = newnode;

}

}

int dequeue(queue **front,queue **rear)

{

int data;

queue *temp;

if(queue_empty(*front,*rear))

return 0;

temp = *front;

data = (*front)->data;

if(*front==*rear)

{

*front = *rear = NULL;

}

else

{

*front = (*front)->next;

}

free(temp);

return data;

}

stack* create_node()

{

stack *newnode;

newnode = (stack*)malloc(sizeof(stack));

newnode->next = NULL;

return newnode;

}

void push(stack **top, int data)

{

stack *newnode;

newnode = create_node();

newnode->data = data;

newnode->next = *top;

*top = newnode;

}

int pop(stack **top)

{

stack *temp;

int data;

temp = *top;

data = (*top)->data;

*top = (*top)->next;

free(temp);

return data;

}

int stack_empty(stack *top)

{

if(top==NULL)

return 1;

else

return 0;

}

void insert(int vi, int vj)

{

adj_list *temp,*newnode;

newnode=(adj_list*)malloc(sizeof(adj_list));

newnode->vertex = vj;

newnode->next=NULL;

if(G[vi]==NULL)

{

G[vi]=newnode;

}

else

{

temp = G[vi];

while(temp->next!=NULL)

{

temp = temp->next;

}

temp->next = newnode;

}

}

void create(int count,char location[MAX][MAX])

{

int i,edges, vi,vj;

for(i=0;i<count;i++)

{

printf("\nEnter the name of node %d\t",i);

flushall();

scanf("%s",location[i]);

G[i] = NULL;

}

printf("\nEnter number of edges\t");

scanf("%d",&edges);

for(i=0;i<edges;i++)

{

printf("\nEnter the edge (node u,node v)\t");

scanf("d",&vi,&vj);

insert(vi,vj);

insert(vi,vj);

}

}

void BFS(int v,int count,char location[MAX][MAX])

{

int w,i,visited[MAX];

queue *front=NULL,*rear=NULL;

adj_list *temp;

for(i=0;i<count;i++)

{

visited[i]=0;

}

enqueue(&front,&rear,v);

printf("\n%s",location[v]);

visited[v]=1;

while(!queue_empty(front,rear))

{

v = dequeue(&front,&rear);

for(temp=G[v];temp!=NULL;temp=temp->next)

{

w = temp->vertex;

if(visited[w]==0)

{

enqueue(&front,&rear,w);

visited[w]=1;

printf("\n%s",location[w]);

}

}

}

}

void DFS(int v,int count,char location[MAX][MAX])

{

adj_list *temp;

int visited[MAX],j,w;

stack *top=NULL;

for(j=0;j<count;j++)

{

visited[j]=0;

}

push(&top,v);

visited[v]=1;

while(!stack_empty(top))

{

w = pop(&top);

printf("\n%s",location[w]);

for(temp=G[w];temp!=NULL;temp=temp->next)

{

if(visited[temp->vertex]==0)

{

push(&top,temp->vertex);

visited[temp->vertex]=1;

}

}

}

}

int search(char location[MAX][MAX], char item[MAX])

{

int i;

for(i=0;i<MAX;i++)

{

if(stricmp(item,location[i])==0) /*string found*/

return i;

}

return -1;

}

void main()

{

int ch,count,i;

char location[MAX][MAX],start[MAX];

do

{

clrscr();

printf("\n*** MENU ***");

printf("\n1.Create");

printf("\n2.Depth First Search");

printf("\n3.Breadth First Search");

printf("\n4.Exit");

printf("\n\nEnter your choice\t");

scanf("%d",&ch);

switch(ch)

{

case 1:

printf("\nEnter number of nodes\t");

scanf("%d",&count);

create(count,location);

break;

case 2:

printf("\nEnter starting place\t");

scanf("%s",start);

i = search(location,start);

if(i==-1)

printf("\nInvalid Location Entered");

else

{

printf("\nDepth First Search...\n");

DFS(i,count,location);

}

break;

case 3:

printf("\nEnter starting place\t");

scanf("%s",start);

i = search(location,start);

if(i==-1)

printf("\nInvalid Location Entered");

else

{

printf("\nBreadth First Search...\n");

BFS(i,count,location);

}

break;

case 4:exit(0);

default:printf("\nPlease enter proper choice!!!");

}

getch();

}while(1);

}

This program performs both DFS and BFS operations.

You can certainly do that ...

printf ("This is a number: 12345\n");

... but that does not have the same value as placing the value in a variable and converting the variable into a string ...

int i = 12345;

printf ("This is a number: %d\n", i);

That's the whole point of format specifiers - to initiate a conversion from one place to another.