only 1 loop will be used, for loop..

[poqwe]

my code sir is

#include<stdio.h>
#include<conio.h>
 void main(){
 clrscr();
 int count=1;
 for(int i=1;i<=8;i++){
 if(count<=8){
 if(i==1){
 printf("%d",i);
 if(i==2);
 else{
 if(i%2==0){
 printf("%d",i); 
 count++;
 if(i==8){
 printf("%d",9-i);
}
}
}
}
}
getch();
}

but the output of that is 1.
what will i do so that the output of this problem is
1468
87654321


Please Help!!

[ne555]

that code doesn't even compile (mismatched braces, main must return int)

#include <iostream>
int main(){
   std::cout << 1468 << '\n' << 87654321 << '\n';
}

[poqwe]

is it cout==printf??? sir the problem here is loop...

[coder777]

The problem is on line 8. Everything after line 8 will only be executed if i==1.

Check your closing braces

[poqwe]

ahh so that's why.. so should i delete it or change ?

#include<stdio.h>
#include<conio.h>
 void main(){
 clrscr();
 int count=1;
 for(int i=1;i<=8;i++){
 if(count<=8){
printf("%d",i);
 if(i==2);
 else{
 if(i%2==0){
 printf("%d",i); 
 count++;
 if(i==8){
 printf("%d",9-i);
}
}
}
}
}
getch();
}

so this will be the result?
ill try it sir thank you

[coder777]

You think to complicated. Use one loop for the first output and a second for the second. Further more better indentation. See:

#include<stdio.h>
#include<conio.h>
void main()
{
  clrscr();
  for(int i=1;i<=8;i++)
  {
    if(i==1) // Note: count<=8 is useless
      printf("%d",i);
    else if(i==2)
      ;
    else if(i%2==0)
      printf("%d",i); 
  }

  printf("\n"); 
  for(int i=1;i<=8;i++)
  {
    // Note: if(i==8){ is useless
    printf("%d",9-i);
  }

  getch();
}

[poqwe]

but sir we only required only 1 loop will be used. so how im i supposed to do that.

[coder777]

we only required only 1 loop will be used.

Ok, then your if cascade makes a [bit] more sense.

The trick is doubling the end value of the loop:

#include<stdio.h>
 int main(){
 int count=8;
 for(int i=1;i<=16;i++){ // Note: 16
 if(i<=8){
 if(i==1)
printf("%d",i);
 if(i==2);
 else{
 if(i%2==0){
 printf("%d",i); 
 }
 }
 }
 else
 {
 if(i==9)
 printf("\n");
 printf("%d",count);
 count--;
}
}
return 0;
}

Without proper indentation!

[poqwe]

but sir, we only use this problem

for(int i=1;i<=8;i++)

no less no more :(( i tried to much but its to complicated

[coder777]

Is it allowed to use an array?

[dhayden]

Please post the full problem. First you just said what the output should be. Then you said it should use a loop. Then you said the bounds of the loop are fixed.

And regarding the output, is it

1468
87654321

or

2468
87654321

The latter would make a whole lot more sense.

[poqwe]

sir i said at first that "This is my CODE" and i post it "the result is or the output of that is "1" then the big real problem here is that , we should only use 1loop and and the result is/ or the output:
1468
87654321

array is not allowed sir :( its to complicated ..
Please Help.

[coder777]

sir i said at first that

Instead you should have posted the requirements given to you.

However. This is the solution:

#include<stdio.h>
 int main(){
 int count=1;
 for(int i=1;i<=8;i++){
 if(count<=8){
 if(count==1)
printf("%d",count);
 if(count==2);
 else{
 if(count%2==0){
 printf("%d",count);
 }
 }
 i--; // Note
 }
 else
 {
 if(count==9)
 printf("\n");
 printf("%d",9-i);
}
 count++;
}
return 0;
}

[WakeofMisery]

http://www.cplusplus.com/doc/tutorial/control/
http://mrbool.com/importance-of-code-indentation/29079

[dhayden]

sir i said at first that "This is my CODE" and i post it

I understand that you posted your code. What I'm asking is for the actual problem, not the program. Something says you can't use an array. Something says you can't use 2 loops. Something says you must use one loop. Something says the loop must go from zero to 8. What is that something? Can you post it?

for(int i=1;i<=8;i++){
    std::cout << 1468 << '\n' << 87654321 << '\n';
    break;
}

Also, are you certain that the first number is 1468 and not 2468?

[poqwe]

The problem sir is what should i do about this problem.. actually we only allowed to use only 1 loop "for loop" and array is not allowed and also the output is

1468
87654321

it makes me sick, we only allowed to use 1 loop "for loop" and "if" "else" only sir and the loop will start from 1 to 8 only..

[coder777]

And none of the provided solutions are according to your taste?

The problem sir is what should i do about this problem.. actually we only allowed to use only 1 loop "for loop" and array is not allowed and also the output is

If that are the requirements dhaydens solution does what you want.

[poqwe]

hmmmm im just asking some knowledge about programming sir... my course is Geodetic Engineering and i cant relate the world of programming to my course.. its to hard for me. sorry sir. if the output of what you given to me is

1468
87654321.. im asking for another so that i can learn from you and from them.. thank you so much sir..