Re: ternary operator

[Steve Graegert <graegerts@gmail.com>]

On 6/28/05, Vadiraj <vadiraj.cs@gmail.com> wrote:
> Hi List,
> 
>  I'm confused with the behavior of this program..
> 
> func4()
> {
>     return 3;
> }
> func3()
> {
>     return 2;
> }
> func2()
> {
> }
> 
> func1()
> {
>     return(func2()?func3():func4()) ;
> }
> int main()
> {
> 
>      printf("%d\n",func1() ) ;                       // this prints 3
>      printf("%d\n",func2()?func3():func4()) ;  // this prints 2
> }
> 
>  I dont understand whats making the two statements to behave differently.

I am not able to recreate this behaviour, since both printf()s give
"2" on my console and this is exactly what I expected to take place. 
Just take a look at our discussion about implicit return values on
int- and void-valued functions a week or two ago.  If no return value
is given explicitly it is undefined and in most cases not zero,
therfore not yielding false.  It's chosen randomly.

In this case the :? operator says:  if func2() returns true, call
func3() else call func4().  In most cases func2() is true.  Nothing
magic here.

-- 
Kind Regards

    \Steve

--

Steve Graegert <graegerts@gmail.com> || <http://www.technologies.de/~sg/>
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