Why my solution doesn’t produce correct output when I = 2?

Question

It should work since the if statement is working for I = 1 and 2.

Problem: https://www.urionlinejudge.com.br/judge/en/problems/view/1098

https://code.sololearn.com/ca22A08A9a21/?ref=app

Answer 1

Alpha Caspian Because decimal is not precise in programming. A famous example is 0.1+0.2 != 0.3. It's not only in C++, but basically every languages which have the same decimal system.

Though float can correctly hold up to 7 decimals, doing calculation with it may produce an approximate or an accurate value. We can't tell which it will be unless output it. Same things happen to double.

By doing printf("%.7f", I), you can see that there is a 2 at the 7th decimal. I guess it's it so 2.0000002 != 2.

That being said, I don't know why in C++ it doesn't output 2.0000002 unless you explicitly set the precision.

If you want, you can use if(I==0 || I==1 || I==2.0000002f). I tested it and it works.

Answer 2

Because this is C++, you can use cout and let it omit the zero decimals output for you.

Remove the if and else statements and add
cout << "I=" << I << " J=" << I+J << "\n";

Answer 3

Alpha Caspian f tells the compiler to treat 2.0000002 as float, without it, the number would be treated as double and the variable I will implicitly convert to double temporarily when compare to equality.

Basically, it won't go well because float can only hold up to 7 decimals, while double is more precise and stores up to 15 decimals. The further decimal won't be zeros but a bunch of inaccurate numbers stored in the float.

There are also L and LL which can be placed after int, indicating long and long long respectively.

Answer 4

Alpha Caspian I don't know, but you can check out the way C++ and most languages store decimal:

https://en.m.wikipedia.org/wiki/IEEE_754#Basic_and_interchange_formats

It's worth to check out, but honestly there isn't a practical use from knowing it in programming as you still cannot prevent the inaccuracy.

Answer 5

.0f part makes it to output decimals

Answer 6

I see. Thank you so much CarrieForle.

Answer 7

CarrieForle I got your point but it should work for printf also with this code. But why it isn’t working for 2? I don't get it.

Answer 8

Alpha Caspian, it does work for I = 2 it outputs 3.0 , 4.0 and 5.0

Answer 9

Aleksei Radchenkov Though it gives the right output but It's not matching the problem requirements here.
You it says
For I=2 j should be 3, 4,5 not 3.0, 4.0 5.0. That means no zero decimals

Answer 10

I is indeed different from 2, it's 2.0000002384...

Everytime you add 0.2 you're introducing in your result a bit of noise
Doing it 5 times is still fine, but 10 times is enough for the error to become noticeable

Answer 11

Alpha Caspian, but that's what you did by printf("I=%.0f J=%.0f\n", I,J+I); which adds 0.0 to output....

Answer 12

But .0f means it will Just print without 0decimals.

Answer 13

Thanks CarrieForle I got your point now.

Answer 14

CarrieForle in the condition u used I==2.0000002f. Is it necessary to put f in last decimay place? Why?

Answer 15

CarrieForle sorry to bother you again. Another question poped on my head. Why the value is increasing in this manner? Like before 2.0000002 it was 1.8000001. I mean at 7th decimal it's increasing. Why?

Answer 16

Ok CarrieForle Again Thanks a bunch.

Answer 17

It's worth to check out, but honestly there isn't a practical use from knowing it in programming as you still cannot prevent the inaccuracy.