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Re: Confused about floats
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Re: Confused about floats


  • Subject: Re: Confused about floats
  • From: Kyle Sluder <email@hidden>
  • Date: Tue, 5 Oct 2010 08:23:39 -0700

On Tue, Oct 5, 2010 at 8:20 AM, Kyle Sluder <email@hidden> wrote:
> The arithmetic conversion rules say that the integer literals will be
> promoted to the appropriate floating point type. So this is perfectly
> fine and does exactly what you expect:

P.S. The promotion doesn't only apply to literals but to other
expressions too. So your second casting example is unnecessary as
well. Compile this with -Wall -Werror to see that is is perfectly
valid C:


#include <stdio.h>
int main(int argc, char **argv) {
  float x = 2.5;
  int y = 5;
  x += y;
  printf("%f\n", x); // prints "7.500000"
  return 0;
}

--Kyle Sluder
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References: 
 >Re: Confused about floats (From: Gregory Weston <email@hidden>)
 >RE: Confused about floats (From: Shawn Bakhtiar <email@hidden>)
 >Re: Confused about floats (From: Scott Ribe <email@hidden>)
 >RE: Confused about floats (From: Shawn Bakhtiar <email@hidden>)
 >Re: Confused about floats (From: Kyle Sluder <email@hidden>)

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