#include <stdio.h> int main() { int n; for(n=7;n!=0;n--) { printf ("n = %d",n--);} getchar(); return 0; }

Question

Can someone please explain why it is running infinitely? I found this somewhere I don't remember. And also I don't know what's this getchar() doing here!

Answer 1

Next time please don't post the code in the question title.

You can paste it in the Code Playground and share the link.

Answer 2

Because in each iteration, the n-- in printf decrement n once, and then after each iteration, the n-- as step also decrement n by one, so each time is decrement 2. So n is 7, 5, 3, 1, -1, -3,... Forever since n!=0 is not met

The getchar() won't get executed as the for loop runs infinitely

Answer 3

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Answer 4

As Gordon up here said n!=0 was not met so it runs forever as your taking 2 away everyone because of n--(you see it 2 times in your code). Also please try at all means to understand, don't rely solely on memory as it always leads to confusion. Then the cycle of you asking these type of questions continue. So take your time and play around with the code then you will understand just by looking :).

Answer 5

4