"Anand J. Bhalerao" <anand@abnar-telsoft.com> writes:

 > Try this sample C++ code on a solaris box.
 >
 >#include <stdio.h>
 >int SetValue(const int &x)
 >{
 >    int &z = (int &)x ;
 >    z = 10 ;

Here you modify an object (the const int y in main)
that was defined as const.  This causes undefined behaviour.
Hence the compiler is allowed to print anything it wants.

 >    printf("address of x = %d - value of x = %d \n", &x, x) ;
 >    printf("address of y = %d - value of z = %d \n", &z, z) ;

This is unrelated to your real question, but here is another
source of undefined behaviour: above you pass an `int *' to
printf("...%d...", ...).  You should either use iostreams instead,
or cast to `(void *)' and use "%p" rather than "%d".

 >    return y ;
 >}
 >int main() {
 >    const int y = 20 ;
 >
 >    printf("address of y = %d - value of y = %d \n", &y, y) ;
 >    SetValue(y) ;

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"Anand J. Bhalerao" <anand@abnar-telsoft.com> writes:
>  Try this sample C++ code on a solaris box.
> int SetValue(const int &x)
> {
>     int &z = (int &)x ;
>     z = 10 ;
...

This is undefined. You have taken an object declared to be const,
cast away its constness, and then modified it. The program may
therefore proceed to do anything at all.
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The answer to your question is in the following line:

>     int &z = (int &)x ;

You're casting away the constness of the integer reference x in assigning
it to z.
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