On 23 April 2017 at 03:59, Andrew McGlashan via luv-main <[email protected]> wrote: > > The return code from ((i++)) operation is different when i has an > initial value of 0.
Well, that is by consistent design, so I wouldn't consider that a bug, myself. Bash provides pre- and post- increment and decrement operators, and you seem to have chosen the wrong one. $ i=0; ((i++)); echo $? 1 $ i=0; ((++i)); echo $? 0 Search 'man bash' for the string "pre-increment". _______________________________________________ luv-main mailing list [email protected] https://lists.luv.asn.au/cgi-bin/mailman/listinfo/luv-main
