logical operator precedence

[huangwason]

To check a file, if it does not exit, assign a value to exit code and print out something, I use the following code:

[ -f $FILE ] || EXIT_CODE=1 && echo "$FILE does not exist"

assure $FILE does not exist, the above logical operator can be regarded as:
(False || TRUE) && TRUE

EXIT_CODE is assigned to 1, it is ok.
but for &&, as long as the return exit code of || is 0, it print out, the result is it always print out, which is not my intension.

is there a solution to make the above logical operation doing like this
False || (TRUE && TRUE)

thanks

 

[feherke]

Hi

Group them :

[ -f $FILE ] || { EXIT_CODE=1; echo "$FILE does not exist"; }

Feherke.

http://rootshell.be/~feherke/

 

[p5wizard]

curly brackets

[tt][ -f $FILE ] || { EXIT_CODE=1 && echo "$FILE does not exist"; }[/tt]

but this is the same:

[tt][ -f $FILE ] || { EXIT_CODE=1; echo "$FILE does not exist"; }[/tt]



HTH,

p5wizard

 

[huangwason]

thanks for your guys quickly reply, it works.

 

[PHV]

I'd use this:

[ ! -f $FILE ] && EXIT_CODE=1 && echo "$FILE does not exist"

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886