Error/Warning - what's "prettier" way of achieving purpose
david_dick at iprimus.com.au
Sun Nov 24 13:34:16 CST 2002
>Some code I've written has following line..
>print (!defined $DEBUG) ? $cgi_int->header : '' ;
>Basically, if I'm debugging, a content type header is printed
>elsewhere.. With the above line I get the following message:
>"Useless use of a constant in void context"
>Should I be using a full-blown if statement?
>Should I be using a different way of saying null?
>Is there a different structure I should be using?
You could use
print ((! defined $DEBUG) ? $cgi->header : '');
print $cgi->header if (not defined $DEBUG); # slightly different, it
won't print the empty string ;)
as to whether you should.... What are you trying to achieve? Quick and
dirty prototyping? Application for others to maintain?
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