How do I iterate through positional variables in bash? - c

Question

The user will give any amount of positional parameters that they want (which are all C programs). I want to make it so that all of the C programs compile. However this is not working; does anyone have a solution?
echo '#!/bin/bash' >> compile
echo if [ "-o"='$1' ] >> compile
echo then >> compile
echo for (i=3; i<='$#'; i++) >> compile
echo do >> compile
echo gcc -o '$2' '${i}' >> compile
echo fi >> compile

Answer 1

Don't use a bunch of echo statements, use a here-doc. Putting quotes around the token after << prevents expanding variables inside the here-doc.
cat <<'EOF' >>compile
'#!/bin/bash'
if [ "-o" = "$1" ]
then
for ((i=3; i <= $#; i++))
do
gcc -o "$2" "${!i}"
done
fi
EOF
Otherwise, you need to escape or quote all the special characters -- you were getting an error because you didn't escape the < in the for() line.
Other mistakes: you need spaces around = in the [ command, and you were missing done at the end of the for loop. And to access a variable indirectly, you need to use ${!var} syntax.
The usual way to iterate over all the arguments is with a simple:
for arg
loop. When there's no in after for variable, it loops over the arguments. You just need to remove the -o outputfile argument first:
output=$2
shift 2 # remove first 2 arguments
for arg
do
gcc -o "$output" "$arg"
done

Answer 2

Here is how I would edit what you originally posted:
$ cat test.sh
echo -e "#!/bin/bash" > compile.sh
echo -e "if [ \"\${1}\" == \"-o\" ]; then" >> compile.sh
echo -e "\tlist_of_arguments=\${#:3} #puts all arguments starting with \$3 into one argument" >> compile.sh
echo -e "\tfor i in \${list_of_arguments}; do" >> compile.sh
echo -e "\t\techo \"gcc \${1} '\${2}' '\${i}'\"" >> compile.sh
echo -e "\tdone" >> compile.sh
echo -e "fi" >> compile.sh
$ ./test.sh
$ cat compile.sh
#!/bin/bash
if [ "${1}" == "-o" ]; then
list_of_arguments=${#:3} #puts all arguments starting with $3 into one argument
for i in ${list_of_arguments}; do
echo "gcc ${1} '${2}' '${i}'"
done
fi
$ chmod +x compile.sh
$ ./compile.sh -o one two three four five
gcc -o 'one' 'two'
gcc -o 'one' 'three'
gcc -o 'one' 'four'
gcc -o 'one' 'five'
for demonstration purposed, I echoed the gcc command in test.sh. To actually run gcc instead of echoing it, change line five in test.sh from:
echo -e "\t\techo \"gcc \${1} '\${2}' '\${i}'\"" >> compile.sh
to
echo -e "\t\tgcc \${1} '\${2}' '\${i}'" >> compile.sh
or pipe the echo to sh like so:
echo -e "\t\techo \"gcc \${1} '\${2}' '\${i}'\" \| sh" >> compile.sh