#include<stdio.h>
#include<conio.h>
void main()
{
int salary[50],n,count=0,result[50];
float HRA,CON,JOLLY,TOT_salary;
clrscr();
printf("enter no. of employees : ");
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
printf("%d) ",i);
scanf("%d",&salary[i-1]);
HRA=0.25*salary;
CON=0.15*salary;
JOLLY=0.10*salary;
TOT_salary=HRA+CON+JOLLY;
printf("\nnet salary for %d) is :%f\n",i,TOT_salary);
if(TOT_salary>10000)
{
count++; // TO COUNT EMPLOYEES GETTING MORE THAN 10,000
result[i-1]=i; //TO SAY WHICH SERIAL NO. GETS ABOVE 10,000
}
}
printf("the total no. of employees crossed rs. 10,000 is :%d & respected numbers as above is :",count);
for(i=0;i<count;i++)
printf("%d)\n ",result[i]);
getch();
}

void main()
{
int COUNT=0,BS,HRA,CA,EA,TOTAL[20] ;
 
FOR(int i=0;i<15;i++)
{
cout<<"enter basic salary";
cin>>BS;
HRA=(BS/100)*25;
CA=(BS/100)*15;

EA=(BS/100)*10;

TOTAL=BS+HRA+CA+EA;
cout<<"total salary of"<<i<< "employee  is"<<TOTAL[i];
}
FOR(INT I=0;I<=14;I++)
{

if(TOTAL[I]>=10000)
COUNT++
}
COUT<<"THE NUMBER OF EMPLOYEES ABOVE 10000 ARE"<<COUNT;
}

C                                                C++
1) C is an structured oriented       object oriented lang.
language
2) C use structures where no       here it contains classes 
function and data's inside       where it contains both data's
 involved . only outside we        as well as member functions
can write functions and 
initilization could be done
3) variables should be           wherever we wann we can 
initilized on the first line       initilize the variable
after the main function

C
  1)Structured programing.
  2)In build functions are used to allocate the memory dynamically.
  3)Struct members are public by default.
  4)Data and functions are separated.

C++
  1)Object oriented programing.
  2)New ,Delete operators are used to allocate the memory dynamically
  3)class members are private by default.  
  4)Together data and functions into single entity.

C
  1)Structured programing.
  2)In build functions are used to allocate the memory 
dynamically.
  3)Struct members are public by default.
  4)Data and functions are separated.

C++
  1)Object oriented programing.
  2)New ,Delete operators are used to allocate the memory 
dynamically
  3)class members are private by default.  
  4)Together data and functions into single entity.

userdefined

it is a  predefined function..............

predfined

The main function is predefine and have three arguments
which return the value to the operating system and main is
also called by operating system .

"userdefined"     because we are writing the code for main()
or we are defining the boby for main()

First we should know whether main() belongs to C/C++ or 
java. If it is in C/C++ it is "user-defined" because not 
available in any header file and body defined by user, but 
compiler will starts body execution from main() function 
only. But in java without main() also we can execute 
programs and get output eg:-using static blocks.
Here also it is "user-defined" one.

main() is a predefined function because compiler starts
executing only in main() function. It better to press F7
function key after compiling your program because the
control first passes only to main() function

#include<stdio.h>
#include<conio.h>
void main()
{
char a[50],temp;
int count=0;
printf("enter the string");
gets(a);
for(int i=0;a[i]!='\0';i++)
count++;
for(i=0;i<count;i++)
{
for(int j=0;j<count;j++)
{
   if(a[j]=a[j+1];
{
  temp=a[j];
 a[j]=a[j+1];
a[j+1]=temp;
}
}
}
printf("\n");
puts(a);
getch();
}

thank u

#include <stdio.h>

int main()
{
  char arr[100];
int count =0;
int j,i,k;
char temp;
printf("enter the string\n");
gets(arr);
for(i=0;arr[i]!='\0';i++)
count++;
printf("value of count is %d\n",count);
for(k=0;k<=count;k++)
{
for(j=0;j<count-1;j++)
{
   if(arr[j]>arr[j+1])
   {
     temp=arr[j];
     arr[j]=arr[j+1];
     arr[j+1]=temp;
   }

}
}
puts(arr);
return 0;
}

This is the correct and working program.

its working but not correct.

Then which one is the answer. i think the above one works
perfectly and they both will yield you the same results
which you wanted.

is this a tricky qustion? if so whats the answer for this

fopen("file1.txt","wb"); will open the file file1.txt in 
binary read and write mode. Binary mode is useful to write 
a file with 0s and 1s. For example, sometimes it would be 
required to open a executable file. In that case, binary 
mode is useful.
fopen("file1.txt","w"); will open the file1.txt in text 
mode which means you can write regular letters and numbers 
in that file.

In structural focus is more on procedures whereas in oject based focus is more on data..

one mr thing.. structural follows top down approach whereas object oriented follows bottom up approach

header file should contain only declarations and structure 
templates.It should not contain function definitions.

Head file should not contain defining instances of global
variables and function bodies.

switch is faster because when in nested if condition has to 
check for each time. where as in switch it diectly check 
only labels.

as for as me is concerned switch is faster....

in if-else first it will check the if condition , if it is 
true it's no problem.. but if it falls false, it will go 
to  the else part ... 

but in switch case , the argument given inside switch 
statement will see and automatically to the necessary case 
of it... so by comaring the time constraint ,switch saves 
the time for checking each else statement for every if....


thank u

switch statement is more faster and consumes less memory 
than if-else statement the reason being that the switch 
statement is applied  when we have a single variable to 
check but in case of if-else different variable may be 
checked at the same time.

According to me, switch is faster. cause in nested if- 
else, the checking occurs in each step. So, more the 
compiler checks, the more it takes time. So switch case is 
faster.

-1 -1 0 2 1

0 0 1 3 1


correct me if im wrong

0 0 1 3 1
 
as for as i know this is the output........

thank u

0 0 1 3 1

0 0 0 3 1

0 0 1 3 1
m=-1&&-1&&0||2;//i.e,m=1;and i,j,m,l are incrimented

0 0 1 3 1 
this answer is write

2222228344

2222228344...

the above code can  be remodified as :
printf("%d",*q); in first printf statement to get the same
output for both printf statements



thank u

output is 2222228344


in loop 1

as we know that array variable contain the base address of 
the array 

*c means we are trying to print the value of contained in 
the base address which is not changed in the loop

note we can't change the base address of the array that is 
we can't do 'c++'

as in loop for 5 times contain of the base address will be 
printed as 
       22222



in loop 2

in assigned base address of the array to pointer p

here we are printing the contain of address stored in p and 
increment the value of p ( that is pointing to the next 
element of the array )

so we will get output for 2nd loop is 

       28344


overall answer is 

2222228344

 if any wroung in my aswer plz info me at

molugu.ashwin@gamil.com

2222228344 will be the answer. I think everybody answered 
before me is correct.

Pointer to constant: If pointer is pointing to constant
variable is caller pointer to constant. We can not change
the value of that constant.
const int constVariable = 6;
int *ptrConstVar = &constVariable;

Constant Pointer: We declare a pointer as constant. We can
change the content pointed by pointer. But we can not do any
airthmatic operation on the pointer like increment or decrement.
int localVariable =10;
const int *p = &localVariable;

we can not do p++ or p--;

Constant pointer :
it is a pointer which points to the same memory location or
to same address and and whatever value the variable which is
pointed by the pointer holds.
for eg :
char a;
char const *p;
p=&a;
here var. a  is a memory location having a address and  that
variable contains some character data . but this pointer        
 p points to the same address ( a ) however the  value in
var. a  changes. finally, THE POINTER POINTED TO AN ADDRESS
IS A CONSTANT ,WHATEVER THE VALUE INSIDE THE VARIABLE MAY BE..

POINTER TO A CONSTANT :
this is a pointer which points to a constant variable
assigned to that pointer. and another pointer can also be
assigned to a same const. variable to point to. 

for eg :
char Const a;
char *p,*q;
p=&a;
q=&a;



thank u

ya, vignesh, your answer is correct. thanx for giving such 
a good answer.

Not Sure..But If the below way is correct. then yes, we can
do that

#include <stdio.h>
int main()
{
  printf("this is first program\n");
  int main()
  { 
     printf("this is second program\n");
     return 0;
  }
 main();
  return 0;
}

I think its not possible.

may be using the preprocessor directives.. like..

#include "prog1.c"
#include<stdio.h>
int main()
{
.
.
.
.
}

You can include a C file in an another file as depicted in the above example. 
And you can also include an another process call ( which is basically a program) in a running program like..

int main()
{
  main() // example 1
  system("./RunMyProg"); // example 2
  system("ls -l"); // example 3
}

12

Output of this program is 22
reason is :
0x10 is hexadecimal value  which is equal to 16 in decimal.
010 is octal value which is equal to 8 in decimal
10 is decimal value 

so total is 16+8+10= 34

I value is 34.
Now we are printing the value of I on hexadecimal using %x
34 is equal to 0x22 in hexadecimal.

so the output is 22 in hexadecimal

Hope its is clear to you

Output of this program is 22
reason is :
0x10 is hexadecimal value  which is equal to 16 in decimal.
010 is octal value which is equal to 8 in decimal
10 is decimal value 

so total is 16+8+10= 34

I value is 34.

Note:The person above me has given the corrst infm however the output is 34.
Try and run the program in C Compiler