Ubuntu – Difference between $’$foo’ and $“$foo”

bashcommand linescripts

I am having difficulties figuring out some the nuances between single and double quotes within a variable context

I define:

foo=pwd

then run these:

echo $'$foo'

Which echos $foo (meaning the first $ in my echo command is dropped)

echo $"$foo"

This echos pwd (which means the bash expands $foo, my variable, to its value)

echo $`$foo`

Finally this echos $~/scripts (I expected it to print ~/scripts and not $~/scripts)

can somebody help me figure out these differences?

Best Answer

echo $'$foo': the $'[...]' token around $foo interprets $foo literally (as $foo) and tries to expand ANSI C-like escape sequences in it, which are not present, so echo yields $foo;
echo $"$foo": the $"[...]" token around $foo expands $foo to its value (pwd) and tries to translate it if the current locale is not POSIX / C; this is not happening, because either the current locale is POSIX / C or a translation for pwd is not available, so echo yields pwd;
echo $`$foo`: the `[...]` token around $foo allows the expansion of $foo, so $foo is expanded to its value (pwd); the expanded value is run in a subshell, whose output (~/scripts) replaces the whole `[...]` token, so echo yields the $ token followed by the ~/scripts token ($~/scripts).

Ultimately, the last one prints $~/scripts because $`foo` is a combination of a literal $ followed by a command substitution; so the leading $ is interpreted as a literal $ and the trailing `$foo` as a command substitution.

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