Use of . / in c++ - operator or float?

Question

The following piece of code outputs 0 and I'm not sure why. I'm not sure what the meaning of of . is in this context. Is it an operator or is it just indicating a float? Is it related to *Int?

#include <iostream>
using namespace std;
int main(){
   int *Int = new int;
   *Int = 1 / 2 * 2 / 1. * 2. / 4 * 4;
   cout << *Int;    
   return 0;
}

You might want to invest in [a couple of decent books](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list/388282#388282) about C++. — Some programmer dude, Sep 13 '20 at 10:55
`2.` is a `double`, just as `2.0` is. A `float` would be `2.f` or `2.0f`. — Jesper Juhl, Sep 13 '20 at 11:19

score 1 · Answer 1 · answered Sep 13 '20 at 10:45

1

It's not an operator. It indicates a double, not a float.

42. means 42.0, and .42 means 0.42. A . alone is a compiler error (rather than 0.0).

If you add a trailing f, it will become a float instead of double, e.g. 1.f, .1f, 1.0f.

answered Sep 13 '20 at 10:45

HolyBlackCat

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score 1 · Answer 2 · answered Sep 13 '20 at 10:46

1

1 / 2 == 0, 0 multiplied by anything is 0 again.

answered Sep 13 '20 at 10:46

Christopher Yeleighton

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Use of . / in c++ - operator or float?

2 Answers2