according to gnu doc,

The backslash retains its special meaning only when followed by one of the following characters: ‘$’, ‘`’, ‘"’, ‘\’, or newline. Within double quotes, backslashes that are followed by one of these characters are removed.

I tried

p="a\$b";echo $p  # a$b
p="a\`b";echo $p  # a`b
p="a\"b";echo $p  # a"b
p="a\\b";echo $p  # a\b

all is ok. But newline??

p="a\newlineb";echo $p  # a\newlineb

What does newline mean in the doc?


Newline is the name of the newline character, i.e. the character usually written as \n in C:

$ p="a\
b"; echo "$p"

The result is ab rather than \a<newline>b since the POSIX standard says:

A <backslash> that is not quoted shall preserve the literal value of the following character, with the exception of a <newline>. If a <newline> follows the <backslash> the shell shall interpret this as line continuation. The <backslash> and <newline> shall be removed before splitting the input into tokens. Since the escaped <newline> is removed entirely from the input and is not replaced by any white space, it cannot serve as a token separator.

But the string is quoted with double quotes... Well, this is a special case:

The <backslash> shall retain its special meaning as an escape character only when followed by one of the following characters when considered special: $, `, ", \, <newline>.

So the double-quoted string means that the backslash-newline is interpreted as if it was not quoted at all, which means it act as a line-continuation and both the backslash and newline are removed.

Note that this happens as you assign to the p variable, not when you echo it.

See also "http://unix.stackexchange.com/questions/171346/security-implications-of-forgetting-to-quote-a-variable-in-bash-posix-shells" about the neccecity of quoting your variables, always.

  • Is the output being shown ab ? Where has the newline vanished when the variable was quoted ? – user218374 Apr 17 '17 at 23:18
  • @RakeshSharma Please read updated answer. – sorontar Apr 19 '17 at 2:22
  • Maybe also highlight the necessity of quoting the echo argument, though it doesn't particularly make a difference in this specific case. – tripleee Apr 19 '17 at 6:10
  • @sorontar Thanks for the edit, but I've replaced it with a POSIX ref instead. :-) – Kusalananda Apr 19 '17 at 6:25
  • @Kusalananda That is perfectly correct and valid (also for bash), so no complaints, instead: Many thanks :). – sorontar Apr 19 '17 at 14:50

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