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0e1f616
(Get the code!)
Oliver, thanks for your investigations. Indeed the break won't be reached, since !(0 && something) is always true.
However, from reading lp.h, it says
#define LP_PERRORP 0x08 /* unchanged input, active low */
so if that bit is zero, it means 'yes', and thus the printer reports an error condition. I'll talk with upstream about this.