How to convert from float to int?

This is a discussion on How to convert from float to int? within the C Programming forums, part of the General Programming Boards category; Hi, this is a really simple problem, but I just can't find a solution. How do you convert a float ...

  1. #1
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    How to convert from float to int?

    Hi,
    this is a really simple problem, but I just can't find a solution.
    How do you convert a float like 12.34 to a rounded integer like 12??

    Thanks to whoever answers, your help is much appreciated

  2. #2
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    If you really want to round the number (i.e. 2.3->2 but 2.6->3) then you need to write your own function. e.g.:
    Code:
    #include <math.h>
    
    int Round(float myfloat)
    {
      double integral;
      float fraction = (float)modf(myfloat, &integral);
    
      if (fraction >= 0.5)
        integral += 1;
      if (fraction <= -0.5)
        integral -= 1;
    
      return (int)integral;
    }
    
    int main(void)
    {
      float myfloat;
      int myint;
      myfloat = 2.3f;
      myint = (int)myfloat;
      printf("%d\n",myint);
      myfloat = 7.6f;
      myint = (int)myfloat;
      printf("%d\n",myint);
      printf("%d\n", Round(2.3f));
      printf("%d\n", Round(7.6f));
      return 0;
    }
    I'm not sure if you strictly need all those casts, but my compiler gives me warnings if I don't include them, and I don't see the harm (maybe someone could tell me if there is a good reason for not explicitly stating casts - thanks!).
    DavT
    -----------------------------------------------

  3. #3
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    thanks guys, very helpful

    thanks thats just what I needed,
    many thanks

  4. #4
    Cat
    Cat is offline
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    Originally posted by DavT
    If you really want to round the number (i.e. 2.3->2 but 2.6->3) then you need to write your own function. e.g.:
    Code:
    #include <math.h>
    
    int Round(float myfloat)
    {
      double integral;
      float fraction = (float)modf(myfloat, &integral);
    
      if (fraction >= 0.5)
        integral += 1;
      if (fraction <= -0.5)
        integral -= 1;
    
      return (int)integral;
    }
    That would work, but this is much faster (no conditionals to slow you down):

    Code:
    int Round(float myfloat)
    {
      return (int)(myFloat + 0.5);
    }
    Works because x rounded to the nearest int is the same as the integer part of x + 0.5.

  5. #5
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    Question

    Isn't there a round function in C99?

  6. #6
    Just Lurking Dave_Sinkula's Avatar
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    >Isn't there a round function in C99?

    Yes.
    7.12.9.6 The round functions

    Synopsis

    1
    #include <math.h>
    double round(double x);
    float roundf(float x);
    long double roundl(long double x);


    Description

    2 The
    round functions round their argument to the nearest integer value in floating-point format, rounding halfway cases away from zero, regardless of the current rounding direction.

    Returns

    3 The
    round functions return the rounded integer value.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  7. #7
    Pursuing knowledge confuted's Avatar
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    Originally posted by DavT
    I'm not sure if you strictly need all those casts, but my compiler gives me warnings if I don't include them, and I don't see the harm (maybe someone could tell me if there is a good reason for not explicitly stating casts - thanks!).
    The only reason not to cast is because it can suppress more warnings than you meant to suppress. I wish I remembered the details/instances of it happening so I could tell you, but I don't.
    Away.

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