CODEWITHSUNDEEP
bashunixsubstitution
SharkTiles
SharkTiles

Reputation: 605

Bash - demostrating the order of substitution

I'm trying to demonstrate the order of substitution; mainly that variable substitution, command substitution and globbing occurs in order one after the other. I executed the following command and I do not get the expected output.

bash-4.1$ a=file*
bash-4.1$ ls $(echo $($(b=$a)))

I expect the output to list all files names beginning with "file", but instead it outputs the list of all files in the directory. Any idea why?

Upvotes: 0

Views: 98

Answers (2)

ormaaj
ormaaj

Reputation: 6577

Parameter and arithmetic expansion, and command substitution are evaluated at the same time, leftmost-innermost to right. Assignments, null and empty expansions, and redirects are all valid simple commands. The assignment is lost to the subshell, and the arguments to echo expand to nothing. echo outputs a newline, but the command substitution strips it, and ls gets no args. Also, if you were expecting a=file* to do something more than assign a literal string, pathname expansion doesn't occur in assignments.

See: http://wiki.bash-hackers.org/syntax/grammar/parser_exec

And a challenge question when you figure all that out. What will be the value of x? (shows more expansion order and some small Bash quirks.)

declare -i x=0
x+=${x:-1} let x+=2 $((x+=4)) {x}<&$((x+=8,0))

Upvotes: 1

Neil
Neil

Reputation: 55392

The $(...) command substitution returns the output of the command, which is blank for an assignment. So you simply end up running ls.

Upvotes: 2

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