Is this how to round?

This is a discussion on Is this how to round? within the C++ Programming forums, part of the General Programming Boards category; Hi everyone, I was just wondering - is this a valid way to round to the nearest integer in C++? ...

  1. #1
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    Is this how to round?

    Hi everyone,

    I was just wondering - is this a valid way to round to the nearest integer in C++?

    Code:
    double round(double x)
    {
    if(x<0){return ceil(x-0.5);}
    else if(x>0){return floor(x+0.5);}
    else {return(x=0);}
    }
    Thanks a lot.

  2. #2
    Registered User whiteflags's Avatar
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    Test your function and see if the answer is right.

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    I tried it and it worked for the numbers I put in, but I didn't know if there was anything I had missed. I guess it's ok!

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    Remember to test the infinities and NaN. Cause right now, NaN will give the wrong answer.
    It is too clear and so it is hard to see.
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  5. #5
    Algorithm Dissector iMalc's Avatar
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    It's valid, but overkill.
    floor rounds towards minus infinity and ceil rounds towards positive infinity. So they both have the same behaviour either side of zero. This means that you don't need to use both. There's also no reason to have a special case for zero because the floor of zero plus 0.5 is still zero. So all you need is:
    Code:
    double round(double x)
    {
        return floor(x+0.5);
    }
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    Thanks for the help - I found iMalc's suggestion in other places, but people seemed to be saying that it didn't work for negatives. On second thought, it does, doesn't it?

    Thanks!

  7. #7
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by george7378 View Post
    Thanks for the help - I found iMalc's suggestion in other places, but people seemed to be saying that it didn't work for negatives. On second thought, it does, doesn't it?

    Thanks!
    It depends whether you want the result to be a double or an integer. It is common to round by casting to int like this:

    Code:
    double x = ...;
    int rounded = (int)(x + 0.5);
    The above does NOT do the right thing if x is negative, because integer truncation goes toward zero, it's not like floor()
    Code:
    //try
    //{
    	if (a) do { f( b); } while(1);
    	else   do { f(!b); } while(1);
    //}

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