Question

Please tell me how to debug like the below statements in Unix.
How to understand what the below code is doing, step by step.

if [ ! -f $R_C_S_L/j_r_d* ]
    then
echo 0>$R_C_S_L/j_r_d
fi

Answer 1

Well you read it just from left to right:

 if [ ! -f $R_C_S_L/j_r_d* ]

I guess if and [ is clear?

 man [

brings man test up, and there we read:

NAME
       test - check file types and compare values

SYNOPSIS
       test EXPRESSION
       test

       [ EXPRESSION ]
       [ ]
       [ OPTION

and further down:

   -f FILE
          FILE exists and is a regular file

which file?

   $R_C_S_L/j_r_d*

is parsed from left, where $ indicates, it is a variable. From writing variables myself, and glueing them together, I know, that it will include only the L, so $R_C_S_L is a variable, then / is the directory delimiter, and j_r_d* is part of a name, with a globbing pattern.

So a directory, referenced by $R_C_S_L, with a file, matching j_r_d* is searched, that’s the test, and ! is the negation, so if the test fails, the inner part is done.

 echo 0 > $R_C_S_L/j_r_d

writes a 0 to the shortest file, matching above pattern, if no such file exists.