for loop condition issue in c++?

Question

I m working with a pattern program but facing one issue my condition is false then the condition is run?

Pattern Programme:

#include<iostream>
using namespace std;
int main()
{
    int i,j,k,n;

    std::cout << "Enter Number:" ;
    cin>>n;

    for(int i=1;i<=n;i++)    //row 
    {
        //first time condition  false 1(k value)<1(i value) not execute first time this is ok
        //second time condition false 2(k value)<2(i value) why this condition run when condition is false????

        for(int k=1;k<i;k++) //space 
        {
            cout << " ";
        }

        for(int j=i;j<=n;j++)  //column
        {
                std::cout << "*" ;
        }
        cout << "\n";
    }
}

execute of program:

ex: user enter 3

first time excite condition properly :

now i=1 and k=1

  for(int i=1;1<=3;i++)    //row 
   {
        for(int k=1;1<1;k++) //space  //1<1 false  ok.
        {
           cout << " ";
        }

Issue with second-time condition:

now i=2 and k=2

  for(int i=1;2<=3;i++)    //row 
   {
        for(int k=1;2<2;k++) //space  //2<2 false  Not Ok problem is here why these condition is run
        {
           cout << " ";
        }

Link Programme:https://onlinegdb.com/Hk9DHuwvL

@DaniloIvanovic why space loop is one-time run that is my issue?? even I write k++ — dips virag
– dips virag, Commented Apr 5, 2020 at 15:25

Jean-Baptiste Yunès · Accepted Answer · 2020-04-05 15:20:49Z

2

when i=2 and k=2, your explanation is wrong, the right one is:

When i=2, inner code of i loop is:

    for(int k=1;k<2;k++) //space, loops one time 
    {
        cout << " ";
    }

    for(int j=2;j<=3;j++)  //column, loops two times
    {
            std::cout << "*" ;
    }
    cout << "\n";

answered Apr 5, 2020 at 15:20

Jean-Baptiste Yunès

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6 Comments

dips virag Over a year ago

why space loop is run one time that is my issue?? even I write k++

john Over a year ago

The space loop runs one time because i == 2. Look at the loop for(int k=1;k<i;k++). If i == 2 then the loop runs one time. How many times were you expecting it to run?

dips virag Over a year ago

problem is that second time i=2 this is right but also k=2 but here k=1 that is the reason still i does solve this thing a

dips virag Over a year ago

I hope my question is understood to you

john Over a year ago

i=2 this is right but also k=2 That is wrong, k = 1 not 2

|

john · Accepted Answer · 2020-04-05 15:36:09Z

0

Here's your loop

for (int k=1;k<i;k++)
{
    cout << " ";
}

This loop starts at 1 and goes up to but not including i. So it runs i - 1 times. If i == 1 then it runs zero times, if i == 2 then it runs one time, if i == 3 then it runs two times, if i == 4 then it runs three times, etc. etc.

You seem to think the the number of times it has run before makes a difference to how many times it runs next, but that is not true. The loop starts again each time it is run.

answered Apr 5, 2020 at 15:36

john

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3 Comments

dips virag Over a year ago

you mean again k value is reset 1 when increasing 2 and now again k value is reset 1 when increased 3 ?

john Over a year ago

@dipsvirag Yes, k is reset each time the loop runs.

dips virag Over a year ago

now my confusion is clearly that is the my doubt thanks john

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