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Groups > gnu.bash.bug > #14327 > unrolled thread
| Started by | Chet Ramey <chet.ramey@case.edu> |
|---|---|
| First post | 2018-07-10 23:17 -0400 |
| Last post | 2018-07-10 23:17 -0400 |
| Articles | 1 — 1 participant |
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Re: Number with sign is read as octal despite a leading 10# Chet Ramey <chet.ramey@case.edu> - 2018-07-10 23:17 -0400
| From | Chet Ramey <chet.ramey@case.edu> |
|---|---|
| Date | 2018-07-10 23:17 -0400 |
| Subject | Re: Number with sign is read as octal despite a leading 10# |
| Message-ID | <mailman.3415.1531279108.1292.bug-bash@gnu.org> |
On 7/10/18 4:57 PM, Isaac Marcos wrote: > Chet Ramey (<chet.ramey@case.edu <mailto:chet.ramey@case.edu>>) wrote: > > On 7/10/18 2:48 PM, Isaac Marcos wrote: > > That is not an integer constant. Integer constants don't begin > with `-'. > > > That makes negative numbers invalid. You need to look at the difference between constants and operators. This is what allows you to write $(( -a )), for instance. > Because of the difference between an operator and a constant. Unary plus > and minus have a higher precedence than arithmetic operators. So if you > expand the `a' to an expression, which is what happens, the expression > consists of an operator (+ or -) and a constant, and that expression has > a higher precedence than the +. You might think about why using `$a' in > place of the `a' would not work all the time. > > > > I don't care. All other shells do this correctly. It makes you the only one > wrong. "All other shells" is kind of a broad statement. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, UTech, CWRU chet@case.edu http://tiswww.cwru.edu/~chet/
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