I really don't have much experience with inline::C, but I took a look at your problem. It appears to me that you really don't want an "always mapping" of undef to NULL pointer because maybe undef in another context might mean a missing int in an array of ints or something else. Here in this particular case, undef means a NULL C pointer to char and I think you have to check for that explicitly.

I wrote some simple test code below. In this case, I used SV* instead of char*. The C code figures out if the SV contains a valid pointer to string or not? I tested with both NULL undef and "asdf" as test inputs - see below. I suspect that something is missing in the case of UTF-8 strings. Also my simple test probably should be more complex because say passing an int would be treated the same as sending undef (the test is just valid string or not?).

Hope that this is on the right track...
Update: I guess the obvious was unsaid: why not use "" instead of undef? A null string is not undef. 

use v5.10;

use Inline 'C';

say "At least we have compiled a bit of C code!";

my $ret= "HD3";

my $output = enctypex_msname("HD2", undef);

say "Output from enctypex_msname() call: '$output', \$ret is '$ret'";

=OUTPUT
CASE 1:
with: my $output = enctypex_msname("HD2", "asdf");
At least we have compiled a bit of C code!
Output from enctypex_msname() call: 'HD2', $ret is 'HD3'
HD2  asdf
name  0000000002D2BC58 retname  0000000002D2BB68

CASE 2:
my $output = enctypex_msname("HD2", undef);
At least we have compiled a bit of C code!
Output from enctypex_msname() call: 'HD2', $ret is 'HD3'
HD2  (null)
name  0000000002C93298 retname  0000000000000000
=cut


__END__
__C__
unsigned char *enctypex_msname(char *name, SV* retname) {
    
   
    char * retname_string = SvPOK(retname) ? SvPV_nolen(retname) : NUL
+L;
    
    printf ("%s  %s\n",name,retname_string);
    
    printf ("%s %p %s %p\n", "name ",name, "retname ",retname_string);
    return name;
}

[download]
PS again on the C code itself:  "static unsigned char    msname[256];"could be trouble. There is only one msname array. If you call this function again, your code will return the same address for msname as it did the last time. Also perhaps if(!name) return(NULL); should be if(! *name) return(NULL); "" does not mean NULL it means pointer to array of char, if first char is \0 then string is "empty" - not the same as a NULL pointer.

In reply to Re: Inline::C and NULL pointers by Marshall
in thread Inline::C and NULL pointers by markong

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