Re: OFF-TOPIC : Confusion about basic C program behaviour

[Learner <ruxyz@yahoo.com>]

Thank you for the useful link .

--- Gerardo García Peña <gerardo@kung-foo.dhs.org>
wrote:
> Learner wrote:
> 
> >Hi,
> >
> >  I know this is not the correct place to post this
> >question, but I feel it would be pretty easy for
> the
> >kernel gurus to explain this.  
> >
> >TEST CODE
> >==========
> >
> >#include <stdio.h>
> >
> >main()
> >{
> >   float a1=1.0;
> >   float a2=2.0;
> >
> >   printf(" \n");
> >   printf(" Size of Float > %d \n", sizeof(float)
> );
> >   printf(" Size of Int   > %d \n", sizeof(int) );
> >   printf(" \n");
> >
> >   printf(" CASE 1 : f-f > %f - %f \n\n", a1, a2);
> >   printf(" CASE 2 : f-d > %f - %d \n\n", a1, a2);
> >   printf(" CASE 3 : d-f > %d - %f \n\n", a1, a2);
> >   printf(" CASE 4 : d-d > %d - %d \n\n", a1, a2);
> >}
> >
> >OUTPUT :
> >
> > Size of Float > 4
> > Size of Int   > 4
> >
> > CASE 1 : f-f > 1.000000 - 2.000000
> >
> > CASE 2 : f-d > 1.000000 - 0
> >
> > CASE 3 : d-f > 0 - 0.000000
> >
> > CASE 4 : d-d > 0 - 1072693248
> >
> >==================================
> >
> >The CASE1 & CASE2 & CASE4 outputs are expected.
> >
> >   Any explanation for CASE3 behaviour. 
> >   I presume it should behave similar to CASE2 and
> >print 0 - 2.000000 .
> >
> >   Could somebody please pin-point what is 
> >lacking in my presumtion .
> >
> >Thanks !
> >
> >  
> >
> I believe this happens because float is promoted to
> double when it is 
> passed as an arg... so you should think that double
> = 8 and int = 4.
> 
> I have been googling and I have found this:
> 
>    
>
http://www-ccs.ucsd.edu/c/function.html#argument%20promotion
>    
> 


		
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