How preserve words containing spaces while expanding variables in bash script?

Question

My scheme is the following:

I have a shell script that executes a command which calls a C program :

name=$1
StringWithSpaces=$name
command="someprogram.out $StringWithSpaces $otherarguments"
$command

where name is a string with spaces s.a. "String With Spaces" passed to the shell from another python script.

My problem is that when I read that argument in C, it is passed as several arguments instead of just one. I have tried $@, $* and all that stuff. I have also tried to make a function in C that separate the several argv[i] within the StringWithSpaces one, but I am a bit stuck. I wish I could read the variable in C just as a single argument, to make the program as simple as I can.

This is the exact shell code (bash):

#!/bin/bash

#$1 Nombre de la base de datos
#$2 $3 gps coordinates
#$4 geoDistance (radio en km)
#$5 imDownload (a 0: solo se descargan GPS, a 1 también imágenes y tags)
#$Disabled keywords (opcional, lista de keywords separados por comas) 

#Generamos el base path
BASE_DIR=`pwd`
#BASE_DIR=${BASE_DIR%/*}
EXE_DIR=$BASE_DIR/FlickrAPI/bin
DB_PATH=$BASE_DIR/data

#Exportamos las librerías, necesario para que funcione el código
export LD_LIBRARY_PATH=$LD_LIBRARY_PATH:${BASE_DIR}/FlickrAPI/lib

cont=1;

#General DB
name=$1
gps="$2 $3"
geoDistance=$4
numImagesDB=3751
ncores=400;
imDownload=$5

dbDir=$DB_PATH/$name
mkdir "$dbDir";

rm -rf "$dbDir"/imagesDB
rm -rf "$dbDir"/tagsDB
rm -rf "$dbDir"/gps.txt

mkdir "$dbDir"/imagesDB
mkdir "$dbDir"/tagsDB
#tidx=`seq 7 $#`;
#keywords="";
#for ((i=7;i<=$#;i++))
#do
#  keywords=$keywords" "${!i};
#done
keywords=$6;
export LD_LIBRARY_PATH=/home/user/anaconda2/lib/
command="$EXE_DIR/get_GPS_bimestral $dbDir $gps z $numImagesDB $ncores $geoDistance $imDownload $keywords"
echo $command
$command

Answer 1

Put the command in an array with:

Command=(someprogram.out "$StringWithSpaces" "$otherarguments")

When array is expanded in quotes using @ to request all its members, as in "${Command[@]}", then bash expands its array member to a single word. So the desired strings with spaces are kept as single strings.

Here is a sample script:

#!/bin/bash -e

function ShowArguments()
{
    for argument in "$@"; do
        echo "Argument:  $argument"
    done
}

Argument1="abc def"
Argument2="pdq xyz"

Command=(ShowArguments "$Argument1" "$Argument2")

echo "${Command[@]}"
"${Command[@]}"

The output from the above is:

ShowArguments abc def pdq xyz
Argument:  abc def
Argument:  pdq xyz

You may want "$otherarguments" to be $otherarguments. Or, if it contains a string with spaces that should be kept as a string, you should handle it the same way, as an array that is expanded with ${otherarguments[@]} in quotes. Here is an example with quoting of one of the variables used to hold arguments:

#!/bin/bash -e

function ShowArguments()
{
    for argument in "$@"; do
        echo "Argument:  $argument"
    done
}

Argument1="Single argument with multiple words"
Argument2=("Multiple argument" with various "numbers of words")

Command=(ShowArguments "$Argument1" "${Argument2[@]}")

echo "${Command[@]}"
"${Command[@]}"

It produces:

ShowArguments Single argument with multiple words Multiple argument with various numbers of words
Argument:  Single argument with multiple words
Argument:  Multiple argument
Argument:  with
Argument:  various
Argument:  numbers of words

After that, you might want to replace echo with a fancier command that quotes its arguments, to show the proper quoting to the user. But that is an aesthetic or user-interface choice; it will not affect the command executed.

Answer 2

My problem is that when I read that argument in C, it is passed as several arguments instead of just one.

in your script replace

command="$EXE_DIR/get_GPS_bimestral $dbDir $gps z $numImagesDB $ncores $geoDistance $imDownload $keywords"
echo $command
$command

by

command="$EXE_DIR/get_GPS_bimestral \"$dbDir $gps z $numImagesDB $ncores $geoDistance $imDownload $keywords\""
echo $command
echo $command | bash

Explanations :

if the shell script b is :

./a.out "$* 3 4"

and c.c is

#include <stdio.h>

int main(int argc, char ** argv)
{
  printf("argc = %d, argv[1] = '%s'\n", argc, argv[1]);
}

then :

pi@raspberrypi:/tmp $ ./b 1 2
argc = 2, argv[1] = '1 2 3 4'

the C program receives only one arg, that is ok

but if the script is modified to be for instance :

command="./a.out \"$* 3 4\""
echo $command
$command

that doesn't work :

pi@raspberrypi:/tmp $ ./b 1 2
./a.out "1 2 3 4"
argc = 5, argv[1] = '"1'

so if you want to store in a var to echo then execute it you can do :

command="./a.out \"$* 3 4\""
echo $command
echo $command | bash

execution:

pi@raspberrypi:/tmp $ ./b 1 2
./a.out "1 2 3 4"
argc = 2, argv[1] = '1 2 3 4'

of course if you want for example 4 received as a separate argument just put it outside the string :

command="./a.out \"$* 3\" 4"
echo $command
echo $command | bash

execution :

pi@raspberrypi:/tmp $ ./b 1 2
./a.out "1 2 3" 4
argc = 3, argv[1] = '1 2 3'