.bash_functions: Calling a function from another one

Question

I have the following two functions defined in my .bash_functions (which gets sourced into .bashrc):

up() {
  if (($# == 0)); then
    cd ..
  else
    for basename; do
      local result=$(_foo)
      echo $result
      cd $result
    done
  fi
}

_foo() {
  echo ${PWD%/$1/*}/$basename
}

While I can execute _foo, when i execute up, up doesn't seem to know _foo. Am I doing something wrong here, or is this just not possible?

Where do you think foo is getting that $1 from? The function gets its own parameters from the call to foo, not some magical global area. — Engineer
– Engineer, Commented May 10, 2024 at 3:11

Karoly Horvath · Accepted Answer · 2012-05-04 11:16:10Z

2

It does "know" _foo, but you don't pass a parameter to _foo, so probably that's causing the confusion.

edited May 4, 2012 at 11:16

answered May 4, 2012 at 9:15

Karoly Horvath

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1 Comment

Dennis Williamson Over a year ago

It sees basename just fine. Variables are global unless declared local. The problem is that _foo() expects an argument ($1) and isn't getting passed one. This is what your example is actually illustrating, by the way.

DerMike · Accepted Answer · 2012-05-04 11:20:05Z

1

Bash scripts are executed sequentially. In your case, _foo() could be defined before up() and everything should work fine.

answered May 4, 2012 at 11:20

DerMike

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1 Comment

Dennis Williamson Over a year ago

It doesn't matter what order they're defined as long as they're called after they're defined. up() { _foo; }; _foo() { echo "bar"; }; up works just fine.

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