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POWhy does \$ reduce to $ inside backquotes [though not inside $(...)]?
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copied!<p>Going over the POSIX standard, I came across another rather technical/pointless question. It <a href="http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_03" rel="nofollow">states</a>:</p> <blockquote> <p>Within the backquoted style of command substitution, <code><backslash></code> shall retain its literal meaning, except when followed by: '<code>$</code>' , '<code>`</code>' , or <code><backslash></code>. </p> </blockquote> <p>It's easy to see why '<code>`</code>' and '<code>\</code>' lose their literal meanings: nested command substitution demands a "different" backquote inside the command substitution, which in turn forces '\' to lose its literal meaning. So, for instance, the following different behavior seems reasonable:</p> <pre><code>$ echo $(echo \\\\) \\ $ echo `echo \\\\` \ </code></pre> <p>But what about '$'? I.e., what's the point or, more concretely, a possible benefit of the following difference?</p> <pre><code>$ echo $(echo \$\$) $$ $ echo `echo \$\$` 4735 </code></pre> <p>As '$' by itself is not ruled out inside backquotes, it looks like you would use either '$' or '\\\$' all the time, but never the middle case '\$'. </p> <p>To recap,</p> <pre><code>$ echo `echo $$` # PID, OK 4735 $ echo `echo \\\$\\\$` # literal "$$", OK $$ $ echo `echo \$\$` # What's the point? 4735 </code></pre> <p>PS: I know this question is rather technical... I myself go for the more modern <code>$(...)</code> substitution all the time, but I'm still curious.</p>

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