bash quotes expansion puzzle

Question

My script looks like:

opts="-x ''"
curl http://somepage $opts

I want the string in $opts appended to the command. I use bash -x test.sh to check the expansion and see that the the single-quotes are removed.

If I change it to:

opts="-x \'\'"

After expansion, there are 4 single-quotes.

Answer 1

What exactly do you see? With the script

opts="-x ''"
echo curl http://somepage $opts
opts="-x \'\'"
echo curl http://somepage $opts

With bash 3.2.39 or 4.1.5, I see

+ opts='-x '\'''\'''
+ echo curl http://somepage -x ''\'''\'''
curl http://somepage -x ''
+ opts='-x \'\''\'\'''
+ echo curl http://somepage -x '\'\''\'\'''

The first call to curl (well, echo curl) has a last argument consisting of two characters ''. The trace escapes special characters: ' appears as '\'' (a common idiom to “escape” single quotes inside single quotes). Formally, ''\'''\''' consists of an empty single-quoted string '' followed by the backslash-quoted character \', then again '', again \', and a final ''. (Ksh shows this as the slightly more readable $'\'\''.) The second call passes four characters \'\'.

Under the normal sh parsing rules, you can't make an empty argument by expanding an unquoted variable. Word splitting cuts only where there's a non-whitespace or quoted character.

Since you're using bash, you can put multiple options in an array. This also works in ksh and zsh.

opts=(-x "")
curl http://somepage "${opts[@]}"

For this particular case, you can override the environment variable instead.

http_proxy= curl http://somepage

Answer 2

This is because what you want is to pass a literal empty string to curl, but what you get is a set of literal quotes since they are already quoted. One reference points to another which simply recommends using a function:

download(){
    curl http://somepage "$@"
}
download -x ''

If you want to observe what your script is actually doing, try running set -x before it.