c – Nested for loop, run series 10000 times

You have an integer overflow. Look, you define f as int

 int i, f = 1;

and f grows fast and became huge, at the end it must have been (19!)^10000 (a quite big 19! raised into 10000 power); however, int value can not be greater than 2^31 - 1, and so the integer overflow comes (you can get negative values, then positive one, zero).

When i == 18, j == 1 it appears, that f == 0 and from now you have

 // when i == 18, j = 1 we have f == 0 due to integer overflow
 calculation += 1.00 / f;

equals to infinity.

Quick amendment is to declare f being float:

 int i; 
 float f = 1.0;

The computation is still ineffcient, but the calculation is now valid.

calculation == 1.7182816 

Edit: note, that f grows fast, at j = 2, i = 8 we have f == infinity. From now on we start doing nothing:

1.00 / infinity == 0

and that’s why

calculation += 1.00 / f;

equals to

calculation += 0;

We can well rewrite

// we don't want add 0 9997 times
for(int j=0; j<=10000; j++) ...

into (all we want is at most 4 outer loops instead of 10000)

for(int j=0; j<=3; j++) ...