Re: Friend operators don't work where class member operators do.
In article <1168331138.569169.17480@11g2000cwr.googlegroups.com>,
"cesar tejeda" <cesarthnews@gmail.com> wrote:
void
incr( int& i )
{
++ i ; // Obviously something more complex in real
// code.
}
void
f()
{
unsigned x = 0 ;
incr( x ) ;
}
In this case, the temporary is the result of an implicit
conversion, and the author is very surprised that the call to
incr doesn't modify x.
So I infer that for making a unsigned to int conversion, C++ creates a
new temporary variable and copies the bits instead of the classical C
behaviour that is doing nothing and understanding the bits of x as bits
representing integers. Am I right?.
No. C doesn't have references. The equivalent C code would roughly be:
void incr(int *i)
{
++*i;
}
void f()
{
unsigned x = 0;
incr(&x);
}
which would usually get you at least a warning about converting an
unsigned int * to an int *.
On the other side, wouldn't it be better not to create temporaries for
the conversion and allow the code you wrote above to be correct and
execute giving correct results?
Well, sure. In general, it is better to have matching types.
--
Nevin ":-)" Liber <mailto:nevin@eviloverlord.com> 773 961-1620
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