NO ERRORS but result...........

if ther s no body for the 'for' loop,it should end with 
semicolon..(i.e)
main()
{
  char c;
  for(c='A';c<='Z';c++);
getch();
}
or
main()
{
  char c;
  for(c='A';c<='Z';c++)
   {}
getch();
}

in this question whenever the loop excuted getch() take 
input after that next loop is excuted and similar getch() 
take another input 
that think do untill the conditions of c<='z' is not 
completed

God help those who help themselves!
try it ur's self because i don't know about this programming
brother! i'm sorry forgive me.

The initial condition is c='A'.
Here the the statement "getch();" is the body of the loop.
it is executed once and c becomes 'B' and again the loop is
executed.
This is done,I mean the loop is executed 26 times.
when c become 'Z' the loop is executed one more time(because
the condition is c<='Z')and then the program exit.

this wont show any errors since syntactically it is correct...
but since there is no instructions after for loop for operations unfortunelety getch() will be taken as the next instruction , and thus compiler system will be under ambiquity. since the character key pressed will be taken as the value of getch() so that will say to the OPERATING SYSTEM that compilation has got over and it will return to the IDE... but for loop has not got over.. so this depends on the compilers usage............



thank u

No Errors
loop will execute for 26 times,each time it will wait for an
input and finally terminate when c becomes > Z.

Void main()
{
   int dec,i=1,rem,res=0;
   Printf("Enter the Value %d",&dec);
   while(dec!=0)
    {
        rem=dec%2;
        dec=dec/2;
        res=res+(i * 1);
        i=i*10;

    }
printf("The Binary value is %d",res);
}

main()
{
    int dec,rem,ans=0;
    printf("Enter the number\n");
    scanf("%d",&dec);
    while(dec>=2)
    {
        rem=dec%2;
        dec=dec/2;
        if(rem==0)
            ans=ans*10;
        else
        ans=(ans*10)+1;
    }
    printf("The binary number is");
    while(ans>0)
    {
        rem=ans%10;
        ans=ans/10;
        printf("%d",rem);
    } 
    getch();
    return 0;
}

void main()
{
   int dec,rem,i=1;
   long int bin=0;
   printf("Enter the decimal number : ");
   scanf("%d",&dec);
   while(dec>0)
   {
      rem=dec%2;
      dec=dec/2;
      bin=bin+(i*rem);
      i=i*10;
   }
   printf("The binary number is %l",bin);
   getch();
}

Explanation:
The output variable bin is taken as long int bcoz it might 
exceed the range of normal int.

e.g.
dec=25

Then

bin=(1*1)+(10*0)+(100*0)+(1000*1)+(10000*1)
   =11001

#include<stdio.h>

int main()
{
int n, rem, num, i=1;

printf("enter no\n");
scanf("%d", &n);

while(n>0 )
 {
    rem = n % 2;
    n = n/2;
    num = (rem * i)+num;
    i = i  * 10;
 }
 
 printf("binary no: %d", num);
}

@sudha
ur code is abs right with a minor mistake.
main()
{
int num,rem,b=0,i=1;
printf("enter num");
scanf("%d",&num);
while(num)
{
          rem=num%2;
          num=num/2;
          b=b+rem*i;
          i=i*10;
           }
           printf("\n%d",b);

}

all these answers crap

when you
b = b+rem*i
b is filled with int number
and adds the next one to it in the loop 
so it wont be as  b= 10011001
it will be b= the number i have intered 
you idiots ....

/* convert a decimal number to binary */
	int dectobin(int dec)
	{
		int bin=0, i=1;
		while(dec!=0)
		{
			bin+=(dec%2)*i;
			dec=dec/2;
			i*=10;
		}
		return bin;
	}

If a float variable multiply by an integer, it will give 
answer in float. So the answer should be in float.

5.0

(a) you can not do 
x=y=z=0.5 in c++ it will throw error..
undefine symbol y & z...

(i++ > j++) gives 0 because 0 > 0 is false so it return 0.
before returning 0 i is 1 ,but it is overwrite by 0.
 In the Conditional operator false means ,it executes i++;
so i is 1.

1.we know that i=j=0 initially

2.then it will checks the non-incremented 'i' value(i.e 0) 
with incremented 'j' value(i.e 1). So obviously condition 
is falls.

3.now false statement has to be executed i.e (i--),before 
executing this 'i' value is (i.e incremented value)'1' 
after executing false condition(i.e i--)the value of 'i' 
becomes  '0'.

4.So the value of 'i' is '0'.

UNIT i,j :
this line indicates that UNIT is an user defined data type. it may been declared as follows :

          typedf int UNIT
we are making the code more readable

i=j=0 :  indicates that the var. i and j are declared as 0

i=(i++>++j) ? i++ : i-- : the process here is 

i++ is an post  incrementation . if this is compared with any relational or any operaters first that value will be operated first and the 'i' will get incremented ........
but ++j if we take first thing it will increment the value and then operation will be performed

so when it is compared first i will be 0 and j will be 1 so 0 is not greater than 1. so false, so it will go to the statement after  ':' so i-- is there so final value of i will be 0.

thank u

total=0;
k=1;
do
{
total=total+k*k*k;
k++;
}while(k<=n)

total=0;
while(n>=1)
{
  total+=n*n*n;
  n--;
}

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
int n=3;
int k=1,total=0;
do
{
    total=total+(pow(k,3));
    k=k+1;
}
while(k<=n);
printf("%d",total);
getch();
return 0;
}

//1st method..
there is no use of j.as explained below
int j,n=psitive value;
do
{
cout<<"*";
n--;
}while(n!=0);
//2nd we can use j..
int j=0;
do
{
cout<<"*";
j++;
}while(j!=n);

total=0;
cin>>amount;
while (amount>=0)
{
total=total+amount;
cin>>amount;
}

total= 0;

scanf("%d",&amount);
total +=amount;
scanf("%d",&amount);
total +=amount;
scanf("%d",&amount);
total +=amount;

Mr.Subbarao

duvvuri subba rao

The question itself is gramatically wrong.........
this is something of the sort "did you eat tomorrow?"

3   3    1

its .....1 2 2

2 2 1

113

1 2 3

the output will be 3 3 1.

sorry for not explaining it.

this is due to a concept of STACK which is a DATA STRUCTURE.

take the statement : printf("%d%d%d",i,++i,i++);

this list of variables will be getting stored in the stack. like the way shown:
        i++
        ++i
         i
since the operation of the stack is  LIFO(last in first out)
the process will be done as said as LIFO but while retriving the data it will be printing according to the printf statement so only the output   3 3 1

331

1 1 2

Yes, but before giving the answer I wanna discuss the question.
In printf() function compiler calculates the values from
right to left (i.e. at first calculates the vale of i++,
then ++i and at last i)but prints the values from left to right.
So compiler at first calculates the value of i++, here i=1
so the value is printed 1 for i++, in the post increment the
value of i becomes 2, but in the pre increment which is ++i,
the value becomes 3, so the value is printed 3 for ++i, now
the value of i is 3, for this reason the value is printed
again 3 for i. But as I said before printf() function prints
from left to right 
so the output will be 3 3 1

Answer is 5..........

declaring int i inside for loop is not available in
traditional 'c'

we can not declare loop condition variable in c..we can do
in c++.

it goes from 0 to 5

 and  print i

why shoul have a cout if is c++

 buth this is c  (printf)

  and why cannot daclare a int in the for ciclus

We can't declare a variable in any part of the program
rather than the declaration part.
It doesn't matter whether you use a loop to print or not.
When the statements which are to be executed begins(Here the
looping statement)no declaration is possible in C.
You can do it in C++,C#,java etc.

n=No of elements in an array;

for(i=0;i<n;i++)
{for(j=0;j<n-i;j++)
   {if(arr[j]>arr[j+1])
     swap(arr[j],arr[j+1]);
   }  
}

void swap(int*x, int*y)
{
  int temp;
   temp = *x;
   *x = *y;
   *y = temp;
}

//n=No of elements in an array;
int n;
int []a;
for(i=0;i<n;i++)
{
 for(j=0;j<n-i;j++)
   {
    if(a[j]>a[j+1])
     {   //swap function for the variables in the array
       int temp=a[j];
        a[j]=a[j+1];
         a[j+1]=temp;
         }
    }
}

//program for selection sort
#include<stdio.h>
#include<conio.h>
#define MX 100
void selection(int [], int);
void selection(int a[],int n)
{
	int minindx,t;
	int i,j;
	for(i=0;i<n-1;i++)
	{
		minindx=i;
		for(j=i+1;i<n;j++)
		{
			if(a[j]<a[minindx])
			minindx=j;
		}
		if(minindx!=i)
		{
		t=a[i];
		a[i]=a[minindx];
		a[minindx]=t;
		}
	}
}
void main()
{
	int a[MX],i,n;
	clrscr();
	printf("enter total num of elements:");
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
	scanf("%d",&a[i]);
	}
	selection(a,n);
	printf("sorted arrau:\n");
	for(i=0;i<n;i++)
	{
		printf("%d \t",a[i]);
	}
	getch();
}