Bash: string of variables as program options

Question

My program takes three options ./program a b c INPUT > OUTPUT (a,b,c are options).
I have a list of option combinations:

cat list_of_combinations
1 2 3
4 8 3 
1 5 7
9 5 6

I want to execute program using those four different combinations. For example:

./program 1 2 3 INPUT >> OUTPUT  
./program 4 8 3 INPUT >> OUTPUT
./program 1 5 7 INPUT >> OUTPUT

I already tried this:

while read COMBO; do
    echo $COMBO | 
     ./program - INPUT >> OUTPUT
done < list_of_combinations

And this:

while read COMBO; do
    ./program $COMBO INPUT >> OUTPUT
done < list_of_combinations

My question is:
How to execute program/command while using a string of variables as options?

Edit

I wouldn't have any problems if there was only one character as variable. What I mean by that:
Program takes only one option:

./program a INPUT > OUTPUT
cat VARIABLES
    1 
    15
    78
while read VARIABLE; do
    ./program $VARIABLE INPUT > OUTPUT
done < VARIABLES

But when there are several options (string of characters), for example ./program a b INPUT > OUTPUT I am getting error.

Your second loop with ./program $COMBO INPUT >> OUTPUT should work as long as the arguments cannot themselves contain spaces. — chepner
– chepner, Commented Oct 21, 2013 at 12:55

anubhava · Accepted Answer · 2013-10-20 16:19:11Z

2

Use quotes around your argument variable:

while read COMBO; do
    ./program "$COMBO" INPUT >> OUTPUT
done < list_of_combinations

Without quotes $COMBO will be considered 3 different inputs rather one single string as with the quotes.

answered Oct 20, 2013 at 16:19

anubhava

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1 Comment

Mark Reed Over a year ago

When @Pgibas said that it should work like ./program 1 2 3 INPUT >> OUTPUT, I assumed he meant that those commands actually worked when typed literally. :\

Mark Reed · Accepted Answer · 2013-10-20 16:10:21Z

1

By putting the options on the command line:

./program $COMBO INPUT >> OUTPUT

So your second example should work. What goes wrong with it?

answered Oct 20, 2013 at 16:10

Mark Reed

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2 Comments

pogibas Over a year ago

I already tried that (check my question). This would work it there was only one character as variable.

Mark Reed Over a year ago

As I said above, what goes wrong? What do you mean "This would work it there was only one character as variable"?

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