Why does'nt ${*// /*} work to replace blanks with * in command line arguments?-CodePudding

I want my script to perform the product of all its integer arguments. Instead of performing a loop I tried to replace blanks with * and then compute the operation. But I got the following result which I don't understand:

#!/bin/bash
# product.sh

echo $(( ${*// /*} ))  # syntax error with ./product.sh 2 3 4
args=$*
echo $(( ${args// /*} ))  # ./product.sh 2 3 4 => outputs 24

How is it that the first one produces an error while using an intermediate variable works fine?

CodePudding user response：

How is it that the first one produces an error:

From the Bash Reference Manual:

If parameter is ‘@’ or ‘*’, the substitution operation is applied to each positional parameter in turn

(emphasis mine)
That is, the expression ${*// /*} replaces spaces inside positional parameters, not the spaces separating positional parameters. That expression expands to 2 3 4 (which gives a syntax error when used in an arithmetic context), since the parameters itself don't contain a space. Try with

./product '2 ' '3 ' 4

and you will see the difference.

CodePudding user response：

You may utilize IFS:

#!/bin/bash
# product.sh

# set IFS to *
local IFS='*'

# use IFS in output
echo "$*"

# perform arithmetic
echo "$(( $* ))";

Output:

2*3*4
24

Reason why echo $(( ${*// /*} )) doesn't work because bash doesn't allow nested expressions.

CodePudding user response：

In your example, the value $* does not actually contain any literal spaces, so ${*// /*} does not do anything.

If it did, those asterisks would be subject to wildcard expansion, so the idea of performing a substitution would seem to be rather brittle even if it worked.

I would simply create a function to process the arguments, instead of rely on trickery with substitutions -- these tend to have icky corner cases when one of the arguments is a variable or etc.

mul () {
    case $# in
      [01]) echo "$@";;
      *) local n=$1; shift; echo $((n * $(mul "$@")));;
    esac
}

CodePudding user response：

Or use printf, like this:

echo $(( $(printf '%s*' $*)1 ))