I have an XS routine that (simplified) looks like this:
foo * routine(a, b, c=NULL) char *a; int b; bar_t * c; CODE: RETVAL = croutine( a, b, (c && SvOK(ST(2)) ) ? c : cdefault ); OUTPUT: RETVAL
From perl, we can say
my $c = $objref; routine( $a, $b, $c);
or
routine( $a, $b);
BUT, if we say:
my $c; $c = $objref if( false ); routine( $a, $b, $c );
We fail with "c is not a reference" at runtime, when the interface code sees that c is not a reference.

This is because $c contains undef. But I'm prepared to supply another value in this case. (Note the SvOK(ST(2))).

How do I tell XS that a reference OR undef is OK for a parameter (c)?

Many thanks,

This communication may not represent my employer's views, if any, on the matters discussed.


In reply to XS: How to enable passing ref or undef? by tlhackque

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