Solved

Nested Loop

Posted on 2001-08-01
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Last Modified: 2011-09-20
#include<iostream.h>
#define VIEW '*'
main()

{     int i,j,k;
     for(i=1; i<=6; i++)
     {
          for (j=1; j<=6-i; j++)
               cout<<" ";
          for (k=1; k<=2*i-1; k++)
               cout<<VIEW;
               cout<<"\n";
     }
return 0;

}

I need to understand this program. Basically the result will appear as pyramid. But how does this program work. I need to know. Can someone explain in depth???

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Question by:chwankok
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Accepted Solution

by:
IainHere earned 10 total points
ID: 6340523
cout << "X";

displays X on the screen.  In this program, there are spaces, stars and newlines being written, in order to build up the pattern you see.  

for(i=1; i<=6 ; i++)
{
...
}

starts off by making i=1, and goes through each of the instructions between { and }.  Then it increases i by one [i++], and continues until i is greater than 6.  From those instructions you should be able to follow through what happens and why the pattern is produced.

A couple of points: the main() function in a C++ program must return an int.  The '\n' is a newline character (makes the following stuff appear on the next line), and should be replaced with endl.  

#define VIEW '*' would be better written as
const char VIEW('*');

#include <iostream.h> is an old version of that header (not guaranteed to exist in C++ in the future), and you should always use
#include <iostream>
which puts everything in the std namespace:

#include<iostream>
const char VIEW('*');

int main()
{    
int i,j,k;
    for(i=1; i<=6; i++)
    {
         for (j=1; j<=6-i; j++)
              std::cout<<" ";
         for (k=1; k<=2*i-1; k++)
              std::cout<<VIEW;
              std::cout<<endl;
    }
return 0;
}
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Expert Comment

by:nietod
ID: 6340535
This is at least a 50 point question....
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Expert Comment

by:IainHere
ID: 6340536
In the other for(;;) loops, because they don't have { } enclosing the following statements, it is only the next statement that is carried out by the loop.

for (k=1; k<=2*i-1; k++)
             std::cout<<VIEW;
             std::cout<<endl;

is therefore the same as

for (k=1; k<=2*i-1; k++)
{
       std::cout<<VIEW;
}
std::cout<<endl;

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Expert Comment

by:Triskelion
ID: 6340568
lainHere is correct, adding the braces in the proper places will help a lot (for clarity) even though the are not absolutely necessary.
Example:
int main(void) /* main function declaration */
{
     int i,j,k;/*the counters are defined here*/
     for(i=1; i<=6; i++)/*count number of rows*/
          {
          for (j=1; j<=6-i; j++)/*number of spaces before the star*/
               {/*loops 5 times, then 4 then 3, etc*/
         cout<<" ";/*prints a space*/
               }
          for (k=1; k<=2*i-1; k++)
               {
               cout<<VIEW; /* prints the star */
               }
          cout<<"\n"; /* prints a carriage return */
          }
     return 0;/* satisfy the return value type of int for main()*/
}
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Expert Comment

by:Axter
ID: 6340723
Hi chwankok,

nietod is right.  This is at least a 50 point question.
Furthermore, I see you have the following grading record:

C A A B A C B C C A

This is a poor grading record.

If you don't have that many points, many experts would still be glad to help you, if they see that you have a good grading record.

When you have a bad grading records, some experts will not help you even if you assign adequate points to the question.

Usually an 'A' grade is given to most questions, unless the expert does not make an honest effort to help you.

Remember that these experts do not get paid for helping you.  The least that you can do to show your gratitude is to give them a decent grade.

Please be considerate of the experts who are trying to help you, and give them the grades that they deserve.

Thank you
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Expert Comment

by:Axter
ID: 6340768
0
 
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Expert Comment

by:The_Brain
ID: 6340815
Well if you ask me he's no older than 14/15 by the looks of his question.  Give him a break -_-

anyway, the thing is simply saying

#include<iostream.h>   //includes input output streams
#define VIEW '*'       //defines VIEW as character *
void main()                 //main procedure
                // (don't forget to say void!!!!! one day you might use int or something for multithreading @.@)
{                          //opening brace so that the compiler knows where to start
                           //it NEEDS NEEDS an opposing brace to signify the end of the body    
 int i,j,k;                //defines variables i j k as integers
    for(i=1; i<=6; i++)    // for loop.  starts i=1 every loop, i increases by 1, (i++ is short for i= i+1)
    {                      //another opening brace for a new body namely the loop
         for (j=1; j<=6-i; j++)   //another for loop, j=1 and increments to till 6-i
              cout<<" ";          //outputs a 'space'
         for (k=1; k<=2*i-1; k++) // another for loop
              cout<<VIEW;         // outputs * remember #define?
              cout<<"\n";         // outputs line break. "\n" moves to next line
    }                             //oposing brace of the first Loop
return 0;                         //not REALLY REALLY needed but extremely good practice to return 0 to a void procedure

}                                 //I'd keep you guessing but this is the main's closing brace :)


now what this loop means


  for(int i=1; i<=6; i++)  //this will loop what EVER IS IN THE BRACES of this LOOP -- 6 times!!
      {  //beginning of the 6 loop
         for (j=1; j<6-i; j++)   this loop in the FIRST time it goes here (when i=1) it will loop 6-1 times (5 for the mathematically challenged)  so it's gonna output 5 spaces =)
cout<<" ";
       }
   next after that loop happened 5 times, this one happens
       for(k=1; k<2*i-1; k++)
         cout<<"*";   //remebmer it's still when i=1 so, this will loop from 1 -> 2*i - 1 ( 1 -> 1)
hence it's gonna only run ONCE and leave
//        so guess what, only gonna output ONE *
          NEXT LINE IS IMPORTANT!!!!!!!!!!! it's NOT NOT NOT NOT NOT NOT PART of the last loop... SO SO many ppl make this mistake! :.(

for EVERYLOOP in GOOD practice say   for(i=something; condition here; increment here)
                                       { <---- ALWAYS even just one line, because ONE day you're gonna add something to your loop where there was only one line, and it's NOT gonna work and you're gonna wonder why, and spend 5 more points askin why  =P lol

and the line break is not part of that last loop and puts a break.


The 'i' Loop runs again the MAIN loop.  And i = 2;  here's a hint, STEP through your code, I hope this helps :p
it took enough time to type this lol ;p
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Expert Comment

by:IainHere
ID: 6340887
The only thing still bothering me is the meaning of these:

@.@
=)
:.(
=P
lol
;p

>>don't forget to say void!!!!!
Surely you mean int?  The standard says int, and sooner or later someone will rely on your program (OK, probably not this one) returning something, and that will cause corruption or confusion.  And you might even find a compiler enforcing it.
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Author Comment

by:chwankok
ID: 6345190
Axter, i have awarded those questions you listed.
Many thanks to IainHere, The_Brain  and Triskelion. Basicailly, i knew the program how it works. As Triskelion highlighted the bracket is important. i have another question

After the First For Loop there is a bracket. What is the use of bracket after the First for Loop. Another word, before the 2nd for loop. It know i will affect this program without this open bracket and close bracket.
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by:nietod
ID: 6345324
the for and while statemetns cause only the next statement after the loop to be executed as part of the loop.  for example

for(i=1; i<=6; i++)
   cout << i;  // this statement is part of the loop.
   cout << endl;  // this statement is not part of the loop.

Note how the first cout statement is part of the loop, since it is right after the condition.  But the 2nd statement is NOT part of the loop.  That "endl" executes only one time  , not 6 times.   The fact that the 2nd cout is indetned is missleading.  it really should be

for(i=1; i<=6; i++)
   cout << i;  // this statement is part of the loop.
cout << endl;  // this statement is not part of the loop.


But what if you have 2 statements you do what to be p[art of the loop?  Well, then you need to make them into one statement.  The { }s do that.  for example

for(i=1; i<=6; i++)
{
   cout << i;  // this statement is part of the loop.
   cout << endl;  // this statement IS part of the loop.
}
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Expert Comment

by:IainHere
ID: 6345383
chwankok,

I don't mean to grumble, but I did answer that very question in my second post.

<tries not to cry>
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by:nietod
ID: 6345472
...and a few more things.  for 5 points.
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Expert Comment

by:Axter
ID: 6345514
>>Axter, i have awarded those questions you listed.
Thank you
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Author Comment

by:chwankok
ID: 6347402
Thank you guys !!
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