Location of cd executable

Question

I read that the executables for the commands issued using exec() calls are supposed to be stored in directories that are part of the PATH variable.

Accordingly, I found the executables for ls, chmod, grep, cat in /bin.

However, I could not find the executable for cd .

Where is it located?

Answer 1

A process can only affect its own working directory. When an executable is executed by the shell it executes as a child process, so a cd executable (if one existed) would change that child process's working directory without affecting the parent process (the shell), hence the cd command must be implemented as a shell built-in that actually executes in the shell's own process.

Answer 2

cd is a shell built-in, unfortunately.

$ type cd
cd is a shell builtin

...from http://www.linuxquestions.org/questions/linux-newbie-8/whereis-cd-sudo-doesnt-find-cd-464767/

But you should be able to get it working with:

sh -c "cd /somedir; do something"

Answer 3

Not all utilities that you can execute at a shell prompt need actually exist as actual executables in the filesystem. They can also be so-called shell built-ins , which means – you guessed it – that they are built into the shell.

The Single Unix Specification does, in general, not specify whether a utility has to be provided as an executable or as a built-in, that is left as a private internal implementation detail to the OS vendor.

The only exceptions are the so-called special built-ins , which must be provided as built-ins, because they affect the behavior of the shell itself in a manner that regular executables (or even regular built-ins) can't (for example set , which sets variables that persist even after set exits). Those special built-ins are:

• break
• :
• continue
• .
• eval
• exec
• exit
• export
• readonly
• return
• set
• shift
• times
• trap
• unset

Note that cd is not on that list, which means that cd is not a special built-in. In fact, according to the specification , it would be perfectly legal to implement cd as a regular executable. It's just not possible, for the reasons given by the other answers.

And if you scroll down to the non-normative section of the specification, ie to the part that is not officially part of the specification but only purely informational, you will find that fact explicitly mentioned:

Since cd affects the current shell execution environment, it is always provided as a shell regular built-in .

So, the specification doesn't require cd to be a built-in, but it's simply impossible to do otherwise.

Note that sometimes utilities are provided both as a built-in and as an executable. A good example is the time utility, which on a typical GNU system is provided both as an executable by the Coreutils package and as a shell regular built-in by Bash. This can lead to confusion, because when you do man time , you get the manpage of the time executable (the time builtin is documented in man builtins ), but when you execute time you get the time built-in, which does not support the same features as the time executable whose manpage you just read. You have to explicitly run /usr/bin/time (or whatever path you installed Coreutils into) to get the executable.

Answer 4

According to this , cd is always a built-in command and never an executable:

Since cd affects the current shell execution environment, it is always provided as a shell regular built-in.

Answer 5

cd is part of the shell; an internal command. There is no binary for it.

Answer 6

The command cd is built-in in your command line shell. It could not affect the working directory of your shell otherwise.

Answer 7

I also searched the executable of "cd" and there is no such. You can work with chdir (pathname) in C, it has the same effect.