Rounding in C

This is a discussion on Rounding in C within the C Programming forums, part of the General Programming Boards category; What is the best method for traditionally rounding (0.5 up and 0.4 down) in C? I have heard adding 0.5 ...

  1. #1
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    Rounding in C

    What is the best method for traditionally rounding (0.5 up and 0.4 down) in C? I have heard adding 0.5 and called floor or subtracting 0.5 and calling ceil but I want this to be conditional where if my user input value is not a whole number then this is function is called.

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    C99 has a function called round() that you can get if you #include <math.h>. Otherwise floor(x+0.5) will also work (note: the ceil technique will round 4.5 -> 4, but it will work for every number strictly greater than a half-integer). You just have to pick one, it doesn't matter. (Note, though, that like most of the math.h functions, round and ceil will both return double rather than int.)

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    So if I specify a variable as an integer and only want to work with whole numbers, what happens if the user inputs 99.7 ?

    If nothing happens it would be safe to say I could round(var); to achieve my desired results?

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    Registered User whiteflags's Avatar
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    Truncation.

    int d = 99.7 results in 99.

    However, you could write a round function easily, say:
    Code:
    #include <math.h>
    int round_to_even ( double d )
    {
        int rv = (int)d;
        if ( (rv &#37; 2) == 0 )
            return rv;
        else
            return 1 + rv;
    }
    Since there are a couple ways to round correctly, actually.
    Last edited by whiteflags; 02-12-2008 at 09:16 PM.

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    Hey thanks.

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    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by blackcell View Post
    So if I specify a variable as an integer and only want to work with whole numbers, what happens if the user inputs 99.7 ?

    If nothing happens it would be safe to say I could round(var); to achieve my desired results?
    You'll also need to be careful how you are inputting your numbers. If you're using %d to read into an int, for example, and the user types 99.7, you're in a lot of trouble. For %d knows what an int is, and therefore when it sees the . it will stop, putting 99 in the int variable and leaving the .7 out to dry (which means there's no way to round to 100 here, since even the fact that there was a decimal is lost). And if you're not careful, every successive call to scanf will then fail, since that .7 will still be there gumming up the works.

    If you need to do input validation, you can search here for it; I seem to remember a couple threads in the not too distant past about how to deal with nefarious users.

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    What you could do is just print the number into a string, then search along for the '.' and look at the number after the '.' Then make the dot a Null (zero) and read the string back into an integerthen increase that number if necessary.
    Might not be to everyones liking as the best way to do it, but at least it works
    And can be done in 2 minutes

  8. #8
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    as simple as it gets:
    Code:
    float f;
    // . . .
    int r = (int)(f + 0.5f);

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    Quote Originally Posted by citizen View Post
    Truncation.

    int d = 99.7 results in 99.

    However, you could write a round function easily, say:
    Code:
    #include <math.h>
    int round_to_even ( double d )
    {
        int rv = (int)d;
        if ( (rv % 2) == 0 )
            return rv;
        else
            return 1 + rv;
    }
    Since there are a couple ways to round correctly, actually.
    Friend will it work for d = 98.7 ?

    i think we should do it as:
    Code:
    #include <math.h>
    int round_to_even ( double d )
    {
        float rv = d - (int)d;
        if ( (rv < 5 )
            return rv;
        else
            return 1 + rv;
    }

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    Registered User whiteflags's Avatar
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    That has problems too. It will return a small number like 8 for most inputs.

    According to wikipedia, this is the round-to-even method:
    • Decide which is the last digit to keep.
    • Increase it by 1 if the next digit is 6 or more, or a 5 followed by one or more non-zero digits.
    • Leave it the same if the next digit is 4 or less
    • Otherwise, if all that follows the last digit is a 5 and possibly trailing zeroes; then change the last digit to the nearest even digit. That is, increase the rounded digit if it is currently odd; leave it if it is already even.

    So it turns out that I did it wrong, but it's easy to do right. You need another variable for the fractional part so you can check that according to the method. And I would probably use the fmod function in math.h to separate a double's fractional part from its whole integer part.

  11. #11
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    Quote Originally Posted by manav View Post
    as simple as it gets:
    Code:
    float f;
    // . . .
    int r = (int)(f + 0.5f);
    Actually, that, as it is written will not work at all.
    Try it! You will see some odd results!!

  12. #12
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    Actually, that, as it is written will not work at all.
    Try it! You will see some odd results!!
    I have tried it and, on the MinGW port of gcc 3.4.5 with Windows XP SP2, there were no odd results as you claimed. Kindly provide your test program, input, output and expected output.

    My own test was:
    Code:
    #include <stdio.h>
    #include <assert.h>
    
    int round_nearest(float f);
    
    int main(void)
    {
        assert(round_nearest(0.0f) == 0);
        assert(round_nearest(0.001f) == 0);
        assert(round_nearest(0.499f) == 0);
        assert(round_nearest(0.5f) == 1);
        assert(round_nearest(0.501f) == 1);
        assert(round_nearest(0.999f) == 1);
        assert(round_nearest(1.0f) == 1);
        assert(round_nearest(1.001f) == 1);
        assert(round_nearest(1.499f) == 1);
        assert(round_nearest(1.5f) == 2);
        assert(round_nearest(1.501f) == 2);
        assert(round_nearest(1.999f) == 2);
        assert(round_nearest(2.0f) == 2);
        assert(round_nearest(2.001f) == 2);
        return 0;
    }
    
    int round_nearest(float f)
    {
        return (int)(f + 0.5f);
    }
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  13. #13
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    Quote Originally Posted by laserlight View Post
    I have tried it and, on the MinGW port of gcc 3.4.5 with Windows XP SP2, there were no odd results as you claimed. Kindly provide your test program, input, output and expected output.

    My own test was:
    Code:
    #include <stdio.h>
    #include <assert.h>
    
    int round_nearest(float f);
    
    int main(void)
    {
        assert(round_nearest(0.0f) == 0);
        assert(round_nearest(0.001f) == 0);
        assert(round_nearest(0.499f) == 0);
        assert(round_nearest(0.5f) == 1);
        assert(round_nearest(0.501f) == 1);
        assert(round_nearest(0.999f) == 1);
        assert(round_nearest(1.0f) == 1);
        assert(round_nearest(1.001f) == 1);
        assert(round_nearest(1.499f) == 1);
        assert(round_nearest(1.5f) == 2);
        assert(round_nearest(1.501f) == 2);
        assert(round_nearest(1.999f) == 2);
        assert(round_nearest(2.0f) == 2);
        assert(round_nearest(2.001f) == 2);
        return 0;
    }
    
    int round_nearest(float f)
    {
        return (int)(f + 0.5f);
    }
    OK maybe it does work, however I was a bit confused about the 0.5f bit, I thought this
    was 0.5*f, which obviously would not work.
    OK so it seems the "f" means it is a float which I can't say I have ever used before.
    So the next question is why do you put an "f" after the numbers?
    It seems to work well without the f? I guess that is what confused me, as I said, initially
    I thought he meant 0.5*f because I don't see why he needs to put an "f" there at all.

    Anyway it does look very confusing to anyone who is used to normal mathematical
    expressions!!

    What is wrong with?
    Code:
    int round_nearest(float f)
    {
        return (int)(f + 0.5);
    }
    Last edited by esbo; 02-16-2008 at 02:45 AM.

  14. #14
    C++ Witch laserlight's Avatar
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    So the next question is why do you put an "f" after the numbers?
    It seems to work well without the f? I guess that is what confused me, as I said, initially
    I thought he meant 0.5*f because I don't see why he needs to put an "f" there at all.
    Floating point literals are of type double unless specified as float by the case-insensitive f suffix. In this case the numbers should be in range of both double and float, so it does not matter either way.
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