sh - POSIX way to find the path of foo without `which`? -


i've encountered machine not have which. how find path executable foo without which in posix compliant shell script, if foo might not present?

the code have following:

  • if which appears available according type which exiting 0, use it
  • otherwise use type foo , depending on type says (does output contain of following: keyword, builtin, alias, hash, function), grab path according position in output.

the main problem this, @chepner , man page point out, type stdout in unspecified format.

my other problem foo might not ever exit, can't execute see happens. want inspect first need know is.

i feel find / -type f -name foo 2>/dev/null take long. suppose iterate on $path find directly. which approach best? iterating on $path or approach in 2 bullets above, or other approach? need solution portable.


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