On 07/10/2012 05:32 PM, Orlob Martin (EXT) wrote:
Hallo,
I was using the build in (( )) function for count up an index. I think I found
a bug in GNU bash, version 4.2.24(1)-release (x86_64-pc-linux-gnu).
((expression)) is one of Compound Commands (not builtin nor function).
Quoting the manual page:
((expression))
The expression is evaluated according to the rules described
below under ARITHMETIC EVALUATION. If the value of the expres‐
sion is non-zero, the return status is 0; otherwise the return
status is 1. This is exactly equivalent to let "expression".
See also a comment below inline.
The (( )) function is working fine, but the corresponding return value does not
match.
Check following small example (have the focus on $?)
----------SCRIPT START-----------------
#!/bin/bash
index=-3
for a in 1 2 3 4 5 6 7 ; do
echo "Index before: $index"
(( index++ ))
echo "Return value: $?"
echo "Index after: $index"
echo "----"
done
----------SCRIPT END------------------
OUTPUT is:
Index before: -3
Return value: 0
Index after: -2
----
Index before: -2
Return value: 0
Index after: -1
----
Index before: -1
Return value: 0
Index after: 0
----
Index before: 0
Return value: 1<-- ?BUG? The index was counted up correctly, from 0 to 1, but return
value of (( ))-function is "1"
No, works like intended. index value is 0 (zero) and thus return value
of (( )) is 1. And the value of index is post-incremented.
Index after: 1
----
Index before: 1
Return value: 0
Index after: 2
----
Index before: 2
Return value: 0
Index after: 3
----
Index before: 3
Return value: 0
Index after: 4
Cheers,
Martin
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Martin Orlob - Systemintegrator
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