OK, short introduction to the intractes of "if":

"if" is followed by a (any) command and it executes first this command, then the "then...."-part if the command returns RC 0 or the "else..."-part, if not. This is always the case. For instance:

Code:

if some_command ; then
     echo "some_command returned 0"
else
     echo "some_command returned something else"
fi

Here, the command "some_command" is executed. If it returns 0 the then-part is executed, if it returns something else the "else"-part is executed.

You may ask yourself where this:

Code:

if [ $? -eq 0 ] ; then

now comes from because there seems no command to be involved: wrong. In fact there is a command involved and it is called "test". "test" does all kinds of tests: integer comparisons, string comparisons and tests for the existence (or certain attributes) of files and it returns a RC of 0 for the comparisons being (logically) true and an RC <> 0 for comparisons being logically false.

Historically, in the first UNIX versions the construct looked like this (and you may see this still in some very old scripts):

Code:

if test $? -eq 0 ; then

or maybe:

Code:

if /usr/bin/test $? -eq 0 ; then

You see, the expression to be evaluated consists just of arguments to "test". "test", is still there, i suggest to read the man page of it.

But because this didn't look like the programming languages the inventors of UNIX knew (they were all programmers, after all), they created a device which made it look more like what they were accustomed to: they created a symbolic link from /usr/bin/test to /usr/bin/[ - in fact "[" is a possible filename - and the line now looked like this:

Code:

if [ $? -eq 0 ; then

In some variants of UNIX (HP-Ux, if i remember right) this link still exists.

That was actually better but it looked like a bracket was opened and never closed. So the final change to /usr/bin/test was that it expected a "]" as the last argument if it was called as "[" - and this created the look we know today:

Code:

if [ $? -eq 0 ] ; then

Notice, that for this reason the following won't work:

Code:

if [ $? -eq 0] ; then

because the last parameter now would be "0]", and the expected final "]" would be missing. Try it, you will see that it leads to a syntax error - for the reasons stated above. The following weird-looking construct, though, works perfectly:

Code:

if [ "$?" "-eq" "0" "]" ; then

The last step in the development was that programmers of UNIX shells (and i think David Korn with his Korn shell was the first) noticed that "test" and its equivalent was called so often that they made it a shell built-in. This will prevent the necessity of loading and starting an external executable (time-consuming activities) and is a speed optimisation. Still, the buil-in "[" works the same as the external /usr/bin/test (again, with the exception of expecting "]" as the last argument) as the man page for the shell will tell you.

I hope this helps.

bakunin