echo from a function
 

I have a function which works fine.
However, if I put it in a test statement then the echo does not work.

I expected to see
"calling my function"
echoed to the screen, when I call it in these two places.
if test myfun
if [ myfun ]

#here is my script
function myfun {
echo "calling my function"
return 1
}

myfun #this works

if test myfun
then
echo one 1
else
echo zero 0
fi

if [ myfun ]
then
echo one 1
else
echo zero 0
fi

test works on the exit code; "$?", like this:

$ myfun
$ echo $?

If $? is 0, then the test worked, otherwise it is a failure.

Suggestion: Say myfun gives 1 as output when it should test positively:

MYFUN=$( myfun )
if [ "$MYFUN" -eq 1 ]
then
echo one 1
else
echo zero 0
fi

The behavior is correct in both cases, but maybe it is not what you want. The first if/then check the exit status of the command
this does not execute the function, but check for the length of the string "myfun". The test command without options is equal to test -n, which checks if the length of STRING is nonzero. Hence the test is true. Maybe you mean this:
this checks the exit code of your function. The second if/then does exactly the same, since [ is a command equivalent to test. In this case you cannot test for the exit code of a command (the correct way is the first one, without square brackets and without test).

No, I mean, when I call the function in the "if [ myfun ] " line, I would expect "calling my function" to go to the screen. Also if I redirect the output from the body of the function, it does not go to output.txt as I would expect.


function myfun {
echo "calling my function"
echo "this will go to output " > output.txt
return 99
}

if [ myfun ]
then
echo one
else
echo zero
fi

I try to be more clear: when you use the syntax
you evaluate an expression, not a command! That is, you don't execute myfun, but test for the literal string "myfun", which is always true! If you want to test a command you have to use this syntax:
without the square brackets. In this case the function myfun is actually executed and the output is displayed in the terminal (or redirected to a file).

All of the posts here are actually correct, but they may appear little confusing. So I will try to sort them out a little.

The functions in bash are handled different from C/C++. Let's say you define
Than if you call it this way: the function is executed and writes "calling my function" to the screen. The return value is stored in variable "$?". Ultimately
will behave exactly as function in C/C++ storing returned value in MYFUN_RET variable.

But be careful.
MYFUN_RET will still hold return value of your function, but otherwise it will behave different. It will store output of "myfun" function to variable MYFUN. So there will be no output to screen. But if you later do you will get "calling my function". In other words construction VARIABLE=function does not store return value into VARIABLE, but output of the function.

Now to that if commands. I think the difference between and is explained quite clearly in the above post. But there is one small trap hidden. When using it does not evaluate the return code of "command" if "command" is your internal function. It will return the value of call to that function.

So if you want a function to do something and than evaluate its return value, the construction should look like

This is not really clear to me. Why it should not evaluate the return code of a function? What is the "value of call to that function"? The return statement causes the function to exit with a specific return code, whereas if omitted, the return code is that of the last command executed inside the function.

Coluxit you are right. I wrote it little bit confusingly. My error.

This will run if return value of "command" is 0. In any other cases it will run That is what I actually meant. The behaviour is quite opposite of what you get in C/C++

thanks
 

Thank you for all those replies. Very good stuff.