What happens if I name a bash script function with the name of a binary located in the PATH?

Question

Suppose I write a function in a bash script, with the name of an available binary, say, pwd:

function pwd(){
    echo '/'
}

Alright, that seems a bit weird, but the question is: what will happen if further in my script I write the commands:

cd /usr
pwd

What pwd will be used? Also, how can I force the use of the other one?

Answer 1

Your function will get called because it hides the pwd builtin.

To force the command to be executed, use the command builtin:

command pwd

From bash manual:

   command [-pVv] command [arg ...]
          Run command with args  suppressing  the  normal  shell  function
          lookup.  Only builtin commands or commands found in the PATH are
          executed.  If the -p option is given, the search for command  is
          performed  using  a default value for PATH that is guaranteed to
          find all of the standard utilities.  If  either  the  -V  or  -v
          option is supplied, a description of command is printed.  The -v
          option causes a single word indicating the command or file  name
          used to invoke command to be displayed; the -V option produces a
          more verbose description.  If the -V or -v option  is  supplied,
          the  exit  status  is  0 if command was found, and 1 if not.  If
          neither option is supplied and an error occurred or command can-
          not  be found, the exit status is 127.  Otherwise, the exit sta-
          tus of the command builtin is the exit status of command.

Answer 2

Function will have precedence. You can check it easily using type pwd.

Considering that pwd is a builtin, you can reach the real implementation using builtin pwd.

If you want to actually get the executable from your system, you can refer to its path, e.g. using $(which pwd).