I'm taking the book "C programming, a modern approach 2nd edtiion" and for the code/exercise below, why exactly the "1" is not being printed?
I get "5 4 3 2" as output, but then why not 1 as well? What am I potentially missing here?
Code:#include <stdio.h> int main (void) { int i, j; for (i = 5, j = i - 1; i > 0, j > 0; --i, j = i - 1) { printf("%d ", i); } return 0; }
The loop ends when either i or j reaches 0. Since j starts 1 lower than i, it reaches 0 before i. Since you are only printing the value of i you see 5, 4, 3, 2.
A little inaccuracy saves tons of explanation. - H.H. Munro
That's it, thank you very much!Originally Posted by john.c
- Please indent your code
- Dry run the code on pen and paper
Your for loop run condition is (i > 0 , j > 0). There is ",", and this case the compiler uses the right-most condition as the loop run condition.
So your loop is the same as:
So, you basically start of with i = 5, j =4.Code:for (i = 5 , j = i - 1; j > 0; --i , j = i - 1)
1st run: i = 5, j = 4
2nd run: i = 4, j = 3
3rd run: i = 3, j = 2
4th run: i = 2, j = 1
Now, i =1 and j = 0 which violates loop condition and you don't get the 1 printed.
Last edited by Zeus_; 11-30-2019 at 01:34 PM.
> The loop ends when either i or j reaches 0. Since j starts 1 lower than i, it reaches 0 before i. Since you are only printing the value of i you see 5, 4, 3, 2.
Isn't the right-most condition taken as your loop condition? I remember reading that somewhere so please correct me if I'm wrong.
Going by what you said, (i > 0 , j > 0) is the same as (i > 0 && j > 0).
@Zeus, you are absolutely correct! I made a mistake there, somehow reading the comma operator as a &&.
But because it's the comma operator, only the second operand (j > 0) is effective.
Thanks for the correction.
A little inaccuracy saves tons of explanation. - H.H. Munro
Haha, you got me bothered for a few moments there lol. I thought I was learning everything wrong the right way
