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Calling an Inner class From outerclass in java

03:33 |

package com.test.Java;

public class A

{

public static void main(String[] args)

{

A a = new A();

A.B b = a.new B();

b.killkarthik();

}

class B

{

private void killkarthik()

{

System.out.println("karthik killed");

}

http://docs.oracle.com/javase/tutorial/java/javaOO/nested.html

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LinkedList

00:55 |

#include <stdio.h>
#include <stdlib.h>
int count;
bool flag=YES;
int choice;
struct object
{
int data;
struct object *address;
};
void printlist(struct object*);
void insert(struct object*,int pos,int data);
void delete(struct object*,int pos);
int main (int argc, const char * argv[])
{

int pos;
int data;
struct object *list;

//create root if the list is empty
if (list==NULL)
{
list = (struct object*) malloc(sizeof(struct object));
list->data = 99;
list->address = NULL;
}

printf("%lu",sizeof(struct object));

while (flag==YES)
{
Printf("\nEnter the choice: 1-insert 2-print 3-delete 5-exit");
scanf("%d",&choice);

switch (choice)
{
case 1:
printlist(list);
Printf("\nEnter the position:\n");
scanf("%d",&pos);
Printf("\nEnter the data:\n");
scanf("%d",&data);
insert(list, pos, data);
break;

case 2:
printlist(list);
break;

case 3:
printlist(list);
Printf("\nEnter the position:\n");
scanf("%d",&pos);
delete(list,pos);
break;
case 5:

Printf("Breaking");
flag = NO;
break;

default:
break;
}
}

return 0;
}
//printing the list and counting the number of objects in them
void printlist(struct object* list)
{
count = 0;

if (list==NULL)
{
Printf("\nlist is null");
}

while (list->address!=NULL)
{
Printf("\n%d",list->data);
list = list -> address;
count++;
}
Printf("\n%d",list->data);
printf("\ncount=%d",count);
}
//inserting elements in their respective position with data
void insert(struct object* list,int pos, int data)
{

if (count==pos-1) //inserting to the last like appending
{
while (list->address!=NULL) {
list = list->address;
}

list->address = (struct object*)malloc(sizeof(struct object));
list->address->data = data;
list->address->address = NULL;
}
else if (pos==1 && count>1)// inserting at first position after header
{
if (list->data == 99)
{
struct object *temp = (struct object*)malloc(sizeof(struct object));
temp->address = list->address;
temp->data = list->data;

list->address = (struct object*)malloc(sizeof(struct object));
list->address->data = data;
list->address->address = temp->address;

free(temp);
}
}
else //else insert it into the middle
{
int i;

if (count>1 && pos >1 && count!=pos-1)
{
while (list->address!=NULL)
{
list = list->address;
i++;
if (i==pos)
{
break;
}

}

struct object *temp = (struct object*)malloc(sizeof(struct object));
temp->address = list->address;
temp->data = list->data;//just to check

list->address = (struct object*)malloc(sizeof(struct object));
list->address->data = data;
list->address->address = temp->address;

free(temp);

}

}

}
void delete(struct object* list,int pos)
{
int localcount;
if (count==pos) //Deleting the last element
{
struct object *temp = (struct object*)malloc(sizeof(struct object));
while (localcount!=count-1)
{
list = list->address;
localcount++;
}

temp->address = list->address->address;
temp->data = list->address->data;

list->address = NULL;

free(temp);

}
else if (pos==1 && count>1)//deleting the first element
{
struct object *temp = (struct object*)malloc(sizeof(struct object));
while (list->address!=NULL)
{
if (list->data == 99)
{
temp->address = list->address;
temp->data = list->data;

list->address = temp->address->address;

free(temp);
break;
}
}

}
else //else deleting the intermediate values
{
int k=0;
struct object *temp = (struct object*)malloc(sizeof(struct object));
if (count>1 && pos >1 && count!=pos-1)
{
while (list->address!=NULL)
{

k++;
if (k==pos)
{
break;
}
list = list->address;
}

temp->address = list->address;
temp->data = list->data;

list->address = temp->address->address;

free(temp);
}

}

}

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Convert any number into roman number

11:08 |

#include <stdio.h>
int numbercount(int);//function to count the number of digits.
int main (int argc, const char * argv[])
{

int number;
printf("\nEnter the number:"); scanf("%d",&number);

//number = numbercount(number);
// for calculating 1000th position
//----------------------------------
int quotient = number/1000;
if (quotient>=1)
{
number = number%1000;
for (int i=1; i<=quotient; i++)
{
printf("m");
}
}

//for calculating 100th position
//-----------------------------------
int hunquotient = number/100;

if (hunquotient>=1)
{

if (hunquotient>=5)
{
printf("d");
hunquotient = hunquotient - 5;
}

number = number%100;
for (int j=1; j<=hunquotient && hunquotient!=0; j++)
{
printf("c");
}
}

//for calculating 10th position
//-----------------------------------

int tenquotient = number/10;

if (tenquotient>=1)
{
if (tenquotient>=5) {
printf("l");
tenquotient = tenquotient-5;
}

number = number%10;
for (int k=1; k<=tenquotient && tenquotient!=0; k++)
{
printf("x");
}
}

//for calculating last digit
//-------------------------------------

if (number>=5) {
number = number - 5;
printf("v");
}
for (int l=1; l<=number; l++) {
printf("i");
}

return 0;
}
//function to count the number of digits
int numbercount(int num)
{
int count = 0;
while (num>0)
{
num = num / 10;
count++;
}

return count;
}

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ArmStrong Number

11:06 |

#include <stdio.h>

void armstrong(void);
int main (int argc, const char * argv[])
{
armstrong();
return 0;
}
void armstrong(void)
{
int remainder,sum,n;

for (int i=1; i<500; i++)
{
sum = 0;
n=i;

while (n>0)
{
remainder = n%10;
sum += remainder*remainder*remainder;
n = n/10;
}
if (sum==i)
{
printf("\n%d",i);
}
}


}

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Recursive and Normal addition

05:48 |

A 5-digit positive integer is entered , write a function to calculate sum of digits of the 5-digit number:

(1) Without using recursion (2) Using recursion

#include <stdio.h>
int add(int);
int recadd(int);
int sum=0;
int r;
int main (int argc, const char * argv[])
{
printf("%d",recadd(12111));
return 0;
}
//without using recursion
int add(int num)
{


while (num>0)
{
r = num%10;
sum = sum +r;
num = num/10;
}

return sum;
}
//using recursion
int recadd(int num)
{
if (num>0)
{
r = num%10;
   sum = sum + r;
recadd(num/10);
}

return sum;

}

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pattern printing2

23:14 |

         1  
       2   3  
     4   5   6  
   7   8   9   10

#include <stdio.h>
int main (int argc, const char * argv[])
{
int print=1;

for (int i=0; i<=3; i++)
{

for (int j=4-i; j>=1; j--)
{
printf(" ");
}


for (int k=0; k<=i; k++)
{
printf(" %d",print);
printf(" ");
printf(" ");
print++;

}
printf("\n");
}


return 0;
}

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Printing pattern

06:40 |

 A B C D E F G F E D C B A
 A B C D E F   F E D C B A
 A B C D E       E D C B A
 A B C D           D C B A
 A B C               C B A
 A B                   B A
 A                       A

#include <stdio.h>
int main (int argc, const char * argv[])
{
int j,k,l;

for (int i=1; i<=7; i++)
{
for (j=65; j<=71-(i-1); j++)
{
printf(" %c",j);
}
for ( k=1; k<=(i-1); k++)
{
   printf(" ");
   printf(" ");
if (k>1)
{
printf(" ");
printf(" ");
}
}
for (l=71-(i-1); l>=65; l--)
{
if (l==71)
{
   continue;
}
printf(" %c",l);
}

printf("\n");
}

return 0;
}

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21 Match stick problem

04:04 |

Write a program for a matchstick game being played between the computer and a user. Your program should ensure that the computer always wins. Rules for the game are as follows:

− There are 21 matchsticks.

− The computer asks the player to pick 1, 2, 3, or 4

matchsticks.

− After the person picks, the computer does its picking.

− Whoever is forced to pick up the last matchstick loses the game.

#include <stdio.h>
int main (int argc, const char * argv[])
{
int matchstick = 21;
int user,computer;

while (matchstick>=1)
{

if (matchstick==1)
{
printf("\nMatch stick status:%d",matchstick);
printf("\nYou loose!!!!!!:(:(");    break;
}

printf("\nMatch stick status:%d",matchstick);
printf("\nEnter the choice (1,2,3,4)):"); scanf("%d",&user);

printf("\nYou picked %d",user);
if (user>=5 || user <=0)
{
printf("\nInvalid value"); continue;
}

computer = 5 - user;

printf("\nComputer picked%d",computer);
matchstick = matchstick - computer - user;

}
}

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Gregorian calendar Date

03:59 |

According to the Gregorian calendar, it was Monday on the date 01/01/1900. If any year is input through the keyboard write a program to find out what is the day on 1st January of this year.

#include <stdio.h>enum days
{
Monday,Tuesday,wednesday,thursday,friday,saturday,sunday
};
int main (int argc, const char * argv[])
{

int sourceyear = 1900;
int currentyear;
printf("\nEnter the currentyear:"); scanf("%d",&currentyear);

int x = ((currentyear -sourceyear )%7);
switch (x)
{
case Monday:
printf("Monday");
break;
case Tuesday:
printf("Tuesday");
break;
case wednesday:
printf("wednesday");
break;
case thursday:
printf("thursday");
break;
case friday:
printf("friday");
case saturday:
printf("saturday");
case sunday:
printf("sunday");
default:
break;
}
return 0;
}

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String functions through pointers

02:44 |

#include <stdio.h>#include <stdlib.h>
int strlength(char *x);
char* strConcatenate(char *,char *);
char* strRev(char *);
int main (int argc, const char * argv[])
{

///////////length///////////////
char *s ="welcome";

char *y = "tocpmging";

printf("Length:%d\n", strlength(s));
///////////concatenation/////////
char *result = strConcatenate(s, y);

   printf("concatenation:%s\n",result);
   free(result);

///////////reversing////////////
char *reverse = "texttrial";

char *resultstring = strRev(reverse);

printf("reversedstring:%s\n",resultstring);
free(resultstring);



return 0;
}
int strlength(char *x)
{
int count=0;
for (int i=0; x[i]!='\0'; i++) {

count ++;

}
return count;
}
char* strRev(char *x)
{
int i,j;
char *s = malloc(strlength(x));

for (i=0;x[i]!='\0'; i++);
j=0;
while (i>0) {

s[j] = x[i-1];

j++;
i--;
}

s[j] = '\0';

return s;

}

char* strConcatenate(char *x,char *y)
{
int i,j;
char *s = (char *)malloc(strlength(x)+strlength(y));


for (i=0; x[i]!='\0'; i++)
{

s[i] = x[i];
}


for (j=0; y[j]!='\0'; j++) {

s[i] = y[j];
i++;
}
s[i]='\0';



return s;

}

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Function pointer and pointer sizes

21:13 |

Here is an example of function pointers and how different pointers remain same in size can be understood through the following code

Just a simple mathematical problem.....of perform the basic functions using function pointers, and also passing functions in void* as a parameter....

#include <stdio.h>
int add(int a,int b);
int sub(int a, void * b);
int mul(int a,int b);
int div(int a,int b);
int operation(int a,int b,int (*function)(int,int));

int main (int argc, const char * argv[])
{
//Every of this address is of 8 bytes

printf("%lu\n",sizeof(int *));
printf("%lu\n",sizeof(float *));
printf("%lu\n",sizeof(void *));
printf("%lu\n",sizeof(&add));



int (*addptr)(int,int);//create a function pointer for add

addptr = &add; //though & is not necessary since its points to the address


printf("%lu\n",sizeof(addptr));


printf("addittion%d\n",(addptr)(2,3));

int (*subptr)(int,void *);//create a function pointer for sub

subptr =⊂

   printf("subtract%d\n",(subptr)(5,(void *)3));//Pass an object of anytype


printf("mutliply->%d\n",operation(2,3,mul));//passing the base address of the method to call

printf("division->%d\n",operation(6,2,div));

return 0;
}
int operation (int a, int b, int (*functocall)(int,int))
{

return (*functocall)(a,b);

}
int div(int a,int b)
{
return (a/b);
}
int mul(int a,int b)
{
return (a*b);
}
int add(int a,int b)
{
return (a+b);
}
int sub(int a,void *b)//resultant is void so can send data of any type
{
return (a-(int)b);
}

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