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Type casting with printf statements on Mac OSX and Linux

I have a piece of code that behaves differently on Mac OSX and Linux (Ubuntu, Fedora, ...). This applies to type casting in arithmetic operations in printf reports. The code is compiled using gcc / g ++.

Following

#include <stdio.h>
int main () {
  float days = (float) (153*86400) / 86400.0;
  printf ("%f\n", days);
  float foo = days / 30.6;
  printf ("%d\n", (int) foo);
  printf ("%d\n", (int) (days / 30.6));
  return 0;
}

created on Linux

153.000000
5
4

and on mac osx

153.000000
5
5

Why?

To my surprise, this works on both Mac OSX and Linux.

printf ("%d\n", (int) (((float)(153 * 86400) / 86400.0) / 30.6));
printf ("%d\n", (int) (153 / 30.6));
printf ("%.16f\n",    (153 / 30.6));

Why? I have no clue. thanks.

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10 answers

try the following:

#include <stdio.h>
int main () {
  float days = (float) (153*86400) / 86400.0;
  printf ("%f\n", days);
  float foo = days / 30.6;
  printf ("%d\n", (int) foo);
  printf ("%d\n", (int) (days / 30.6));
  printf ("%d\n", (int) (float)(days / 30.6));
  return 0;
}

Notice what is going on? The culprit is double floating point conversion. Remember that float is always converted to double in the varargs function. I'm not sure why poppy will be different. Better (or worse) IEEE arithmetic implementation?

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#include <iostream>
int main () {
  float days = (float) (153*86400) / 86400.0;
  std::cout << days << std::endl;
  float foo = days / 30.6;
  std::cout << (int) foo << std::endl;
  std::cout << (int) (days / 30.6) << std::endl;
  return 0;
}

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#include <stdio.h>

int main(void)
{
    volatile double d_30_6 = 30.6;
    volatile float f_30_6 = 30.6;

    printf("%.16f\n", 153 / 30.6);   // compile-time
    printf("%.16f\n", 153 / d_30_6); // double precision
    printf("%.16f\n", 153 / f_30_6); // single precision

    return 0;
}

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