Round() Function

This is a discussion on Round() Function within the C Programming forums, part of the General Programming Boards category; Hi All, I am trying to use round() function but gcc couldn't find it... Can you please help me?? Compiler: ...

  1. #1
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    Round() Function

    Hi All,
    I am trying to use round() function but gcc couldn't find it...
    Can you please help me??

    Compiler: gcc 4.1.2
    Command: gcc round_func.c

    Code:
    round_func.c
    #include <stdio.h>
    #include <math.h>
    
    int main(void)
    {
      double a = 13.05;    
      printf("round of  %.1lf is  %.1lf\n", a, round(a));
      return 0;
    }
    Error Message:
    round_func.c:7: warning: incompatible implicit declaration of built-in function ‘round’
    /tmp/cckuuW3l.o: In function `main':
    round_func.c.text+0x24): undefined reference to `round'



    Thanks and Regrads,
    Dharak

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    There is no round() function in the C standard library.

    floor() (which rounds down) and ceil() (which rounds up) are both standard, and can be used to implement other styles of rounding (round nearest, round toward zero, round away from zero, ..... ).
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy in reply to you, it is likely you deserve it. Suck it up, sunshine, and read this, this, and this before posting again.

  3. #3
    DRK
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    Because there's no such a function. Use floor() or ceil() instead.

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    Quote Originally Posted by DRK View Post
    Because there's no such a function. Use floor() or ceil() instead.
    Code:
        #include<stdio.h>
        #include<math.h>
    
        int main()
        {
              double a, b, c ;
              a = 3.8;
              b = floor(a);  //Not Working
              c = floor(104.54) //Working
            printf ("Floor of %0lf is %01f\n", a, c);
            return 0;
        }
    Here if I pass constant value in floor function, then its working but if I pass variable then it is not working.

    any idea?

    Do I need to include something else also?

    Thanks and Regards,
    Dharak

  5. #5
    C++ Witch laserlight's Avatar
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    How does it not work?
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    I also dont understand, gcc version is 4.1.2

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    C++ Witch laserlight's Avatar
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    That wouldn't make sense since the result is the third argument to printf, i.e., c, not a, should be changed to b, if necessary.

    EDIT:
    Quote Originally Posted by Dharak Shah
    I also dont understand
    What I mean is: what is it about the results that you get that makes you think that it doesn't work? For example, in the code that you posted, the obvious reason why it appears not to work is that your input value for the variable is different from the constant.
    Last edited by laserlight; 04-15-2013 at 04:50 AM.
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    Quote Originally Posted by ZuK View Post
    You should output
    Code:
    printf ("Floor of %0lf is %01f\n", b, c); // a is not rounded
    Kurt
    That's not a issue. If you remove print statement then also it won't work with variable.

  9. #9
    ZuK
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    Code:
    #include<stdio.h>
    #include<math.h>
     
    int main()
    {
          double a, b, c ;
          a = 3.8d;
          b = floor(a);  //Not Working
          c = floor(104.54); //Working
        printf ("Floor of %0lf is %0lf\n", a, b);
        return 0;
    }
    my output
    Code:
    Floor of 3.800000 is 3.000000

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    Your posted code doesn't even compile because you are missing a semicolon on line 9.

    And we can't help you if you don't tell us what doesn't work. When I run a corrected version of your code, everything works as expected.

    So what's your problem?

    Bye, Andreas

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    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Dharak Shah
    That's not a issue. If you remove print statement then also it won't work with variable.
    It works. It works for me, and ZuK just posted that it works. Therefore, you are either not telling the truth, or you are simply mistaken. I prefer to think the latter. So... the ball is in your court. What is the program that you compiled and ran? What output did you get, and what makes you think that it doesn't work?
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    My Program is:
    Code:
    #include<stdio.h>
    #include<math.h>
      
    int main()
    {
          double a, b;
          a = 3.8;
          b = floor(a); 
        printf ("Floor of %0lf is %0lf\n", a, b);
        return 0;
    }
    and now output is

    Code:
    [sdharak@Quartz Final]$ gcc round_func.c
    /tmp/ccl2Wy6h.o: In function `main':
    round_func.c:(.text+0x24): undefined reference to `floor'
    collect2: ld returned 1 exit status
    gcc version is
    gcc -dumpversion
    4.1.2

    Now I dont understand what is the issue with my setup?
    It should work..


    Thanks,
    Dharak

  13. #13
    C++ Witch laserlight's Avatar
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    You probably need to explicitly link to the math library with the -lm option.
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  14. #14
    ZuK
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    Quote Originally Posted by laserlight View Post
    You probably need to explicitly link to the math library with the -lm option.
    I can see the problem now.
    If you just use
    Code:
    c = floor(104.54);
    it is not necessary to link the mah library. If you use a variable the math lib is needed.
    I don't have an explanation for that, other then gcc is smart enough to evaluate the floor of a literal at compile time.

    Kurt
    Last edited by ZuK; 04-15-2013 at 05:15 AM.

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    Quote Originally Posted by laserlight View Post
    You probably need to explicitly link to the math library with the -lm option.
    Ohhhh...
    It's working... Thanks....
    But why I need to do this?
    Is there any issue with gcc version??

    -Dharak

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