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[DIFF] 15:34 [INFO] GreyCat [1]
pool-173-61-210-188 [2]
#01 Revert to revision 459. Completely incorrect assertion. Try this: var="-n"; echo "$var" Even if your assertion WERE true for bash >= n we would still have to mark it a pitfall for other shells/versions.
#02 Removed claim that echo can't safely display a variable like var="-n Text". As of bash 4.1.7, echo $var _will_ interpret the leading -n as a flag, but echo "$var" will print exactly what's expected.
[DIFF] 07:13 [INFO] BinaryZebra Correct quoting of echo's
[DIFF] 22:24 [INFO] ip32-43-211-87
[DIFF] 13:21 [INFO] StephaneChazelas a few corrections and improvements.
[DIFF] 14:00 [INFO] GreyCat [1]
semmlers123 [2]
#01 Revert to revision 376. spam
[DIFF] 12:17 [INFO] izabera improved example
[DIFF] 20:41 [INFO] GreyCat
[DIFF] 10:59 [INFO] AnthonyGeoghegan [1-2] #02 Add paragraph and example on the exit status of a command substitution


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