./a.out && echo $? only prints return value if return = 0

Question

echo $? in the following scenario does not display the return value, i.e. 1.

echo "int main() { return 1; }" > foo.c
gcc foo.c && ./a.out && echo $?

However, echo $? works when the program returns 0. Why is this?

Note: If you do another echo $? after the code above, you get the desired output 1.

Because the semantics of `&&` is to only execute the second command if the first command returned 0. — celtschk, Sep 15 '14 at 20:15
If you want to run your `a.out` program and then print the return value (a.k.a. exit status) unconditionally, use `./a.out; echo $?`. — G-Man Says 'Reinstate Monica', Sep 15 '14 at 20:17
I would up vote but I don't have enough reps, thanks for the explanation and suggestion! — Harvinder, Sep 15 '14 at 20:22

score 1 · Accepted Answer · answered Sep 16 '14 at 14:31

The && operator is a boolean and with short-circuit evaluation. This means that it only executes the second command if the first one is successful (i.e. it returns 0). A typical way it's used is something like this:

tar xf file.tar && rm file.tar

This only removes the file if the extraction is successful.

Your script also contains a good example of this:

gcc foo.c && ./a.out

will only try to run the program if gcc was successful.

If you want to display $? regardless of the success of a.out, you can write:

gcc foo.c && { ./a.out ; echo $? ; }

The {...} groups the command so they'll both be executed if the compilation is successful.

./a.out && echo $? only prints return value if return = 0

1 Answers1