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From: scjones@thor.UUCP (Larry Jones)
Newsgroups: comp.lang.c
Subject: Re: Nasty bug
Message-ID: <147@thor.UUCP>
Date: 31 Aug 90 13:24:37 GMT
References: <0093BF08.7F3834E0@rigel.efd.lth.se>
Organization: SDRC, Cincinnati
Lines: 33

In article <0093BF08.7F3834E0@rigel.efd.lth.se>, e89hse@rigel.efd.lth.se writes:
> 
> prnval(s,f)
> char   *s;
> float   f;
> {
>    if(f == 0.0)
>        sscanf(s,"%f",&f);
>    printf("%10.2f\n",f);
> }
> 
> And it didn't work. Why? The answer is that the parameter f is a
> double, not a float since all floats are converted to double when they are
> passed as arguments to functions. Therefore &f is a ptr to double rather than a
> ptr to float as one would expect looking at the declartion.

It is worth noting that different C compilers handle this situation
differently.  Some do exactly what you said, they rewrite your
parameter declaration as double rather than float.  Others assign
the passed double value to a local float which is then used as the
parameter.  If the float and double formats are sufficiently similar,
the compiler can just ignore the second half of the passed double
value and leave the parameter a float without having to make a copy.

ANSI C does not allow the declaration rewriting -- it requires the
compiler to make the parameter have the type you declared it as.
So, this problem should not be a problem much longer, right?
----
Larry Jones                         UUCP: uunet!sdrc!thor!scjones
SDRC                                      scjones@thor.UUCP
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