Why does the following command not insert new lines in the generated file?

$ echo "Line 1\r\nLine2" >> readme.txt

$ cat readme.txt
Line 1\r\nLine2
Best Answer


echo

An echo implementation which strictly conforms to the Single Unix Specification will add newlines if you do.

echo 'line1\nline2'

But that's not a trustworthy behavior In fact, there really isn't any standard behavior which you can expect of echo .

OPERANDS

string

A string to be written to standard output. If the first operand is -n, or if any of the operands contain a <\backslash> character, the results are implementation-defined.

On XSI-conformant systems, if the first operand is -n, it shall be treated as a string, not an option. The following character sequences shall be recognized on XSI-conformant systems within any of the arguments:

\a - Write an <alert>.

\b - Write a <backspace>.

\c - Suppress the <newline> that otherwise follows the final argument in the output. All characters following the \c in the arguments shall be ignored.

\f - Write a <form-feed>.

\n - Write a <newline>.

\r - Write a <carriage-return>.

\t - Write a <tab>.

\v - Write a <vertical-tab>.

\\ - Write a <backslash> character.

\0num - Write an 8-bit value that is the zero, one, two, or three-digit octal number num.

And so there really isn't any general way to know how to write a newline with echo , except that you can generally rely on just doing echo to do so.

A bash shell typically does not conform to the specification, and handles the -n and other options, but even that is uncertain. You can do that

shopt -s xpg_echo
echo hey\\nthere

hey
there

And not even that is necessary if bash has been built with the build-time option...

--enable-xpg-echo-default

Make the echo builtin expand backslash-escaped characters by default, without requiring the -e option. This sets the default value of the xpg_echo shell option to on, which makes the Bash echo behave more like the version specified in the Single Unix Specification, version 3. See Bash Builtins, for a description of the escape sequences that echo recognizes.

printf

On the other hand, printf 's behavior is pretty tame in comparison.

RATIONALE

The printf utility was added to provide functionality that has historically been provided by echo. However, due to irreconcilable differences in the various versions of echo extant, the version has few special features, leaving those to this new printf utility, which is based on one in the Ninth Edition system.

The EXTENDED DESCRIPTION section almost exactly matches the printf() function in the ISO C standard, although it is described in terms of the file format notation in XBD File Format Notation.

It handles format strings which describe its arguments - which can be any number of things, but for strings are pretty much either %b yte strings or literal %s trings. Other than the %f ormats in the first argument, it behaves most like a %b yte string argument, except that it doesn't handle the \c escape.

printf ^%b%s$ '\n' '\n' '\t' '\t'

^
\n$^    \t$

See Why is printf better than echo ? for more.

echo() printf

You might write your own standards conformant echo like...

echo(){
    printf '%b ' "$@\n\c"
}

...which should pretty much always do the right thing automatically.

Actually, no... That prints a literal \n at the tail of the arguments if the last argument ends in an odd number of <backslashes> .

It doesn't happen

echo()
    case    ${IFS- } in
    (\ *)   printf  %b\\n "$*";;
    (*)     IFS=\ $IFS
            printf  %b\\n "$*"
            IFS=${IFS#?}
    esac