opposite return status in c shell


 
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# 1  
opposite return status in c shell

there is something wrong with my system. when I do this:
Code:
diff file1 file1 && echo 1

the output is 1.

but
Code:
diff file1 file2 >/dev/null && echo 1

output nothing

while
Code:
diff file1 file2 >/dev/null || echo 1

shows 1.

the same with "grep" return status. they are both GNU utilities.

What could be so messed up to cause this opposite return status? (the return status was correct has been correct until today.)

Thanks.


Moderator's Comments:
Mod Comment Please use code tags when posting data and code samples!

Last edited by Franklin52; 01-14-2011 at 05:43 AM..
# 2  
In many systems, including shell script, it does "lazy" evaluation in "if" statements. If the first expression is "true" and there is a logical "and" it will go onto the second. If it is "false" the "and" statement will immediately be false and therefore will not run the second expression. In your example if the "diff" returns "false" it will not run the "echo".
diff file1 file1 is returning "true" and will run the echo. In the second example diff is returning "false" and therefore not running the echo. In the third example diff is returning "false" but the expression is "or" so it will run the echo.
I hope this (rather verbose) answer is helpful....
# 3  
Hi Citaylor:

I totally understand that part.

But for GNU diff, the return status of "diff file1 file1" should be 0 (no difference found), while "diff file1 file2" should be 1.

Therefore, the "diff file1 file1 && echo 1" should not print out anything. But in our system, it print out 1.

so basically, the return status of diff (or grep) is opposite to typical.

"man diff" shows "An exit status of 0 means no differences were found, 1 means some differences were found, and 2 means trouble."

Does your system behave the same way?

Thanks.
# 4  
Im afraid that is incorrect. A return code of zero in UNIX land means "true" and one means "false". Try running:
Code:
true ; echo $?
false ; echo $?

This User Gave Thanks to citaylor For This Post:
# 5  
@citaylor

Yes, shell return status is "opposite of typical". It is a consequence of using a exit status that is limited to a single value between 0 and 255. Indicating why a command fails with multiple and different 0s is a little difficult Smilie, so instead 0 is used to indicate success and non-zero indicates failures.

<imho>I think it would have been better to use two status variables -- something (aka $?) to indicate success or failure and something else ($: perhaps) to indicate WHY it failed. I'll just need to find a WABAC machine and talk to Mr. Thompson, Ritchie, et al.</imho>
This User Gave Thanks to m.d.ludwig For This Post:
# 6  
Thank you very much for the answers. It makes sense now.

But, does it mean "0" is considered "true" as the return status, but is "false" as a value?

for example:
Code:
set tmp=0;  if ( $tmp ) echo t

will not echo t.

Thanks.

Moderator's Comments:
Mod Comment Please use code tags when posting data and code samples!

Last edited by Franklin52; 01-14-2011 at 05:44 AM..
 

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