Why my solution doesn’t produce correct output when I = 2? | Sololearn: Learn to code for FREE!

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

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

8/3/2021 12:01:46 PM

Samia Haque

18 Answers

New Answer

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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.

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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";

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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.

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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.

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.0f part makes it to output decimals

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I see. Thank you so much CarrieForle.

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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.

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

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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

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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

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

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

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Thanks CarrieForle I got your point now.

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

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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?

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Ok CarrieForle Again Thanks a bunch.

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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.

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