if statement and calling function in if using bash

Question

I have written a function:

check_log(){     if [ -f "/usr/apps/appcheck.log" ]     then          return 1     else          return 0     fi } 

Then I call this function in an "if" condition:

if [ check_log ]; then     ........statements.... fi 

Will this work? I am confused here because bash returns 0 on success and 1 on failure, but my function is returning 1 and the condition is checking for 1/0, it gets 1 and it should give failures, but in my shell script the condition is passing.

Can anyone shed light on this issue?

Answer 1

if [ check_log ]; 

When you use square brackets you're invoking the test command. It's equivalent to if test check_log which is shorthand for if test -n check_log, which in turn means "if "check_log" is not an empty string". It doesn't call your check_log function at all.

Change it to this:

if check_log; 

By the way, the function could be more simply written as:

check_log() {     ! [ -f "/usr/apps/appcheck.log" ] } 

The return value from a function is the exit status of the last command, so no need for explicit return statements.