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Comment by ribcage | Hacker News Reader

Yes. The famous echo on Linux systems does not have it and therefore it's impossible to print the string "-n o p e", because -n will be interpreted as an option.

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echo is not portable anyway, use "printf %s STRING" or "printf '%s\n' STRING".

Yes, that's what I use. Sometimes I still get tempted to use echo because there's less typing...

It does if single or double quotes are used, right? Which would be necessary (or preferred to multiple backslashes) quite often.

No, the quotes are not seen by the program. The program receives a list of strings, it does not get the information about whether and how those strings were originally quoted in the shell. Programs can also be directly called with lists of strings as in execve, so often it does not even make sense to ask if the arguments were quoted or not.

Quotes live on a different level of abstraction.

> No, the quotes are not seen by the program. The program receives a list of strings, it does not get the information about whether and how those strings were originally quoted in the shell.

With quotes the program will receive a single argument -n␣o␣p␣e instead of multiple ones -n, o, p, e. At least it works on the machine here:

    ]$ echo "-n o p e"
    -n o p e
    
    ]$ /bin/echo "-n o p e"
    -n o p e

Yes, I think there was some misremembering here. The nontrivial thing is to print out -n itself with echo. For example, echo doesn't treat "--" specially, so "echo -- -n" prints "-- -n".

Note that this is true for POSIX sytems but not e.g. for Windows. There the program receives the command-line as-is and is responsible for parsing it into an array. There are two different standard functions to do this parsing for you (with slightly different quoting behavior) but you could also create your own that requires options to not be quoted.

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