asked on

Nested Loop

#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???

ASKER CERTIFIED SOLUTION

IainHere

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nietod

This is at least a 50 point question....

IainHere

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;

Triskelion

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()*/
}

Axter

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

Axter

The_Brain

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

IainHere

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.

chwankok

ASKER

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.

nietod

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.
}

IainHere

chwankok,

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

<tries not to cry>

nietod

...and a few more things. for 5 points.

Axter

>>Axter, i have awarded those questions you listed.
Thank you

chwankok

ASKER

Thank you guys !!