# Exrcicses 3 on control flow

I really don’t understand exrcicses 3 on control flow, the code says o - 6 because i is 0 at first and 6 is n. But then there is only 5 spaces printed out? why?

Hi @A_Huang, Thanks for reaching out to us.
The following `for` loop and variable `j` is responsible for printing the space:

``````for (j=0; j<(n-i); j++)
{
printf(" ");
}
``````

Now let’s consider the 6th iteration and value of n=6 when space is not printed

`j<(n-i)` -----> `6< 6-6` -----> `6<0` -----> False

The spaces are not printed at the 6th iteration because the condition `j<(n-i)` in the for loop is false and `print(" ")` not executes at this time.

Hope it will clear your confusion, Happy Learning

So it’s 0< 6, 1 < 6, 2 <6, 3 < 6, 4 < 6, 5 < 6, 6 <6(false). But then there are still 6? I’m happy to see replys but I still don’t understand

1 Like

` for (j=0; j<(n-i); j++) { printf(" "); }`

Hi @A_Huang, Thanks for reaching out. Each line of the pattern carries spaces, if we consider the first line there are 5 spaces.
Condition for first-line: `0<6`, ––> 5 spaces
Condition for second-line: `0<5`, ––> 4 spaces
Condition for third-line:`0<4`, ––> 3 spaces
Condition for fourth-line:`0<3`, ––> 2 spaces
Condition for fifth-line:`0<2`, ––> 1 spaces
Condition for sixth-line:`0<1`, ––> 0 spaces
0<0 ––> Condition false