How does the escape character '\' work in bash prompt?

Question

I started tweaking around with my bash prompt lately and I find myself not understanding how the escape character works. I have the following:

PS1="\[$RED\]\342\224\214\342\224\200"

In this I get it, \[ escapes the [ character and \xxx escapes my UTF-8 characters. But in the following line I get a weird result:

PS1+="$([·\$?·!=·0·]·&&·echo·\[$RED\]\342\234\227\·)"

This will always print X in my prompt, yet if I escape the first $ it will print it only when exit status of any command is non zero. I do not understand why. Wasn't $(commands) supposed to output the result of given commands? If I escape it like so \$() is the whole sequence escaped or just the dollar sign? If I don't escape why doesn't it print $? It just prints the X. I have the same question for the $ inside the square brackets. Why do I have to escape it?

Also I believe this qualifies as another question but is there any way of printing the actual exit status in my prompt?

Answer 1

In a double-quoted string, command substitutions ($(...)) and variable expansions ($foo) are processed, and the backslash in front of the dollar sign prevents that, removing the backslash. This happens during the assignment PS1="$(...)" or PS1="\$(...)".

But the same expansions are also processed when the prompt is printed, so if the dollar sign was escaped on assignment, the resulting PS1 has an unescaped dollar sign, and the expansion happens when the prompt is printed.

So, with an unescaped command substitution, the command only runs once, when the prompt is assigned. With the backslash, it runs every time the prompt is printed.

The difference should be easy to test with these two:

PS1="$(date) "
PS1="\$(date) "

The \[ sequence is different, though. It's only relevant when the prompt is processed, not in regular double-quoted strings. It's used to mark parts of the prompt that don't print any visible characters. Moreover, it only works in the prompt before expansions, so something like PS1='$(echo "\[...\]")' will likely not do what you want.