Re: Calling a function with a float argument modifies caller's state.

[David Brown <david at westcontrol dot com>]

On 22/08/2012 15:58, Jeff B wrote:
> > > On 08/21/2012 06:40 PM, Jeff B wrote:
> > >
> > >
> > > (int)floatVarA = 0.000000            <<=== initial value 0??
> > On 08/22/2012 03:50 AM, Andrew Haley wrote:
> >
> > The initial value is zero because there is garbage in Float Register 0.
> >
> > > floatVarA = 1234.000000
> > Now Float Register 0 contains 1234.0 .
> >
> > > (int)floatVarA = 1234.000000     <<=== changed
> > Which gets printed out.
> >
> > > floatVarB = 5678.000000
> > Now Float Register 0 contains 5678.0 .
> >
> > > (int)floatVarA = 5678.000000    <<=== changed again
> > Which gets printed out.
> >
> > > floatVarA = 1234.000000
> > etc...
> >
> > Moral: undefined behaviour can do anything.
> >
> > Andrew.
>
>
> OK, then I guess the presence of the cast makes it OK for the compiler
> to pull the bits for a variable from somewhere other than where the
> variable actually is.
>
> So, while we are at it, what other casts should I avoid?
>
> Thanks.

You are still missing the point.

Avoid writing code with undefined behaviour.  Do not lie to your 
compiler.  Do not tell it "here is a float" and pass it an integer.  Use 
"-Wall" to let the compiler help you avoid such mistakes.

It is perfectly okay to cast a float to an int - but it is not okay to 
do so and then expect the compiler to treat it as a float.  So you can 
write this:

	printf("(int) floatVarA = %d\n", (int) floatVarA);

Then you are casting the float to an int legally (1234.0 will turn into 
1234), then passing that int to printf, with a format string that tells 
printf to expect an int and print it out.