In bash, how can I directly return the return value returned by calling a function (sry, I don't know how to express this better). Example:
foo() {
echo "$1"
return 1
}
bar() {
return foo 1
}
bar
If I do this, bash complains that a numeric parameter for the return statement is needed.
EDIT
I updated my example to better express the real problem. No only do I want to return the return code, I also want to pass a value to the function first... not sure if this is actually doable.
The only thing you can return from a shell script or a shell function is a numeric error code.
However, you can print some text to standard output in the function (or separate script, it's the same) using echo, cat, etc., and then capture the output, using bacticks syntax or $(...) syntax. Passing parameters to shell functions works the same way as passing parameters to scripts:
In these modifications of your example, I change the argument to foo to make it easier to distinguish the result of one from the other.
foo() {
echo "$1"
return 1
}
bar() {
return "$(foo 2)"
}
bar
echo "$?"
The preceding will output "2". The echo in foo is used as the return value of bar. The range of values that return (and exit) can handle is 0 to 255.
foo() {
echo "$1"
return 1
}
bar() {
foo 2
return "$?"
}
bar
echo "$?"
The second version will first output 2 since that's what foo does then a 1 will be output since that's the return value of bar having been propagated from the return value of foo.
"$(foo 2)" will execute foo in a subshell which is not what I want.
This should work: you can only return a number in bash.
foo() {
return 1
}
bar() {
foo
return 1
}
bar
You can simply return the return code of the last call $?:
foo() {
echo "$1"
return 1
}
bar() {
foo "bla"
return $?
}
bar
f() { g; }; g() { false; }; f; echo $?. Only if there were more commands would you have to save the exit code and return it explicitly.