bash echo command -e arguments ignored using echo full path

Question

What's happening here :

$ echo $SHELL
/bin/bash
$ echo -e "foo"
foo
$ which echo
/bin/echo
$ /bin/echo -e "foo"
-e foo

The last output is expected to be "foo", like when using echo directly.

Answer 1

/bin/echo is a program, and it's different from the Bash built-in echo command.

According to the POSIX standard, echo should not take any arguments. The Bash echo command doesn't follow this standard, and the GNU Coreutils version doesn't either. Other versions may be more strict.

Try to find out what version /bin/echo is by calling it with the --version flag or see if it gives you some help with the --help flag.

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Also, man which shows this:

which returns the pathnames of the files (or links) which would be executed in the current environment, had its arguments been given as commands in a strictly POSIX-conformant shell.

Bash is not strictly POSIX-conformant so the return value of which may differ (and it does in this case because of the echo built-in).

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For further reading, see the answers to echo outputs -e parameter in bash scripts. How can I prevent this?