Why is my program called "set" not being executed?

Question

I've created a simple C program like so:

int main(int argc, char *argv[]) {

    if (argc != 5) {
       fputs("Not enough arguments!\n", stderr);
       exit(EXIT_FAILURE);
    }

And I have my PATH modified in etc/bash.bashrc like so:

PATH=.:$PATH

I've saved this program as set.c and am compiling it with

gcc -o set set.c

in the folder

~/Programming/so

However, when I call

set 2 3

nothing happens. There is no text that appears.

Calling

./set 2 3

gives the expected result

I've never had a problem with PATH before and

which set

returns ./set. So it seems the PATH is the correct one. What's is happening?

Answer 1

Instead of using which, which doesn't work when you need it most, use type to determine what will run when you type a command:

$ which set
./set
$ type set
set is a shell builtin

The shell always looks for builtins before searching the $PATH, so setting $PATH doesn't help here.

It would be best to rename your executable to something else, but if your assignment requires the program to be named set, you can use a shell function:

$ function set { ./set; }
$ type set
set is a function
set ()
{
    ./set
}

(That works in bash, but other shells like ksh may not allow it. See mikeserv's answer for a more portable solution.)

Now typing set will run the function named "set", which executes ./set. GNU bash looks for functions before looking for builtins, and it looks for builtins before searching the $PATH. The section named "COMMAND EXECUTION" in the bash man page gives more information on this.

See also the documentation on builtin and command: help builtin and help command.

Answer 2

set is a builtin in bash (and probably most other shells). This means that bash will not even search the path when looking for the function.

As a side remark, I would strongly advise against adding . to the path for security reasons. Imagine for example cding out of /tmp after any other user added an executable file /tmp/cd.

Answer 3

set isn't just a builtin, it is a POSIX special builtin. There are a few builtin commands which are standards-specified to be found in a command search before anything else - $PATH is not searched, function names are not searched, and etc. Most builtins which are not special are actually required by the POSIX standard to be found in your $PATH before the shell will run any of its own builtin procedures. This is true of echo and most others (though whether the standard is honored in this respect has been a matter of contention at the open group mailing lists in the past), but not of set, trap, break, return, continue, ., :, times, eval, exit, export, readonly, unset, or exec.

All of these are reserved names of the shell, and they have special attributes other than their order of preference for command search as well. For example, you cannot define a shell function with any of those names in a standards-compliant shell. This is a good thing - it enables people to write portable scripts safely. These are baseline commands from which an experienced script-writer can establish a secure and reliable foothold in his or her environment. Invading this namespace is not advisable.

However, if you do desire to invade it, you can do so portably with alias. The order of shell expansion enables this work-around. Because the alias is expanded while the command is being read, whatever you replace the set name with in your definition will expand correctly, it just probably should not expand to one of those names.

So you could do:

alias set=./set

...which will work just fine.

Answer 4

The problem is that set is a shell builtin and the best solution would be to use a different name for your executable program.

Incidentally, last week, I asked a question on how to run system commands instead of shell builtins with the same name and the solution I accepted was to run the command through env:

env set 2 3

For this particular case, where you already know that the command you want to use is located in your current directory, it would be better to directly run the executable by entering its path (using . to represent the current working directory):

./set 2 3

Both the above solutions are shell agnostic, i.e., they will work regardless of which shell you’re using.

Suggestions such as using the command builtin won’t work in Bash: this only prevents shell functions from being run. While, it’s not documented, I’ve also noticed that using command also suppresses shell keywords. However, it won’t do the same for shell builtins such as set. As I understand it, command may work with other shells such as zsh.

Also, tricks such as \set or "set" or 'set' don’t work for Bash builtins – though they are useful for running executable files instead of aliases or shell keywords.

Note: This answer originally began as a comment on Eric’s (accepted) answer but grew too big to fit into a comment. The other answers recommending the use of type and not adding . to the PATH are good ones.