Can you use variables inside $'...' in bash?

Question

I have tried the following:

myprogram $'Hello $HOME'

But it did not work (the command line argument received by myprogram was Hello $HOME, and not the value of $HOME).

Answer 1

Apart from interpreting the C-style backslash-escapes, $'...' works like a single-quoted string. At least Bash's manual mentions this, right at the end of the page on "ANSI-C quoting":

The expanded result is single-quoted, as if the dollar sign had not been present.

So, no. That's probably why they chose to use single-quotes for that (or the other way around), though I don't know about the history of the feature.

You'll have to use double-quotes, and change the quotes as necessary. i.e.

$ myprogram "Hello $HOME"

or with a tab mixed in

$ myprogram $'Hello\t'"$HOME"

Answer 2

You have to write it as

$ myprogram "Hello $HOME"

Bash will preserve the literal meaning of anything enclosed within double quotes except $,`(back-tick) or .So Here it considers "Hello" as the value while $Hello preserves its value as a variable.

Answer 3

Apart from interpreting the C-style backslash-escapes, $'...' works like a single-quoted string. At least Bash's manual mentions this, right at the end of the page on "ANSI-C quoting":

The expanded result is single-quoted, as if the dollar sign had not been present.

So, no. That's probably why they chose to use single-quotes for that (or the other way around), though I don't know about the history of the feature.

You'll have to use double-quotes, and change the quotes as necessary. i.e.

$ myprogram "Hello $HOME"

or with a tab mixed in

$ myprogram $'Hello\t'"$HOME"

Answer 4

You have to write it as

$ myprogram "Hello $HOME"

Bash will preserve the literal meaning of anything enclosed within double quotes except $,`(back-tick) or .So Here it considers "Hello" as the value while $Hello preserves its value as a variable.