Makefile和命令行行为不同

Question

I am currently working on a bigger project where i want to test executable file with few different codes as input.
I call it like this ./test < code1 and after command echo $? , it shows last returned value [0, 1, 2, ..]

I wanted to automate is, so i created call in makefile like this :

#makefile 
[...]
test : 
    ./test < code1 
    @echo $$?
    ./test < code2 
    @echo $$?
    [...]

[...]

So i can call make test .

When program returns 0 as success, everything works fine. But when program has to return something else than 0, it shows me this :

./test < code3
Makefile:19: recipe for target 'test' failed
make: *** [test[ Error 2

Weird thing is, when i try to call program with code which made it crash in command line like :

./test < code3; echo $?  

It works perfectly and shows me last exit status ( for exapmle 3 ).
I am confused now, because i thought it should work the same. Can someone help me out?
Thank you!

Answer 1

See this answer: https://stackoverflow.com/a/41452754/939557

You need to put the echo into the same logical line as your test invocation:

test : 
        ./test < code1; echo $$?
        ./test < code2; echo $$?