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I came across the following line of code (source):

IFS=$'\r'

I'm not quite sure how to interpret that line (specifically why there is a $ character before the newline). It seems like the "special variable" named IFS is being set to a variable named "the newline character"?

What does this line do, and what part of Bash allows this?

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IFS=$'\r' set IFS variable to carriage return.

bash allows ANSI-C Quoting string. $'string' will expands to string, with backslash-escaped characters replaced as specified by the ANSI C standard.

  • How is that different from just doing IFS="\r"? – IQAndreas Apr 4 '15 at 16:45
  • @iqa IFS="\r" sets IFS to the two-character string backslash, r. IFS=$'\r' sets it to a one-character string, a carriage return. – Gilles Apr 4 '15 at 22:41
  • @Gilles: oh, I only think in case double quotes, maybe I'm too sleep, thanks. – cuonglm Apr 5 '15 at 1:55
  • @Gilles Argh, I keep mentally picturing strings in Bash the way they are treated in ECMA-based languages. I need to stop doing that. – IQAndreas Apr 5 '15 at 5:29

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