Problems with eval function skipping newlines in $(eval echo $myvar) - bash

Question

I am currently working on a script meant to add a new project generation for any basic editor.
I am using the following structure in order to generate the right basic program (hello, world) according to the language selected by the user :
#!/bin/sh
#this is a short example in the case the user selected C as the language
TXTMAIN="\$TXTMAIN_C"
$TXTMAIN_C="#include <stdlib.h>
#include <stdio.h>
int main(int argc, char const* argv[])
{
printf(\"hello, world\n\");
return EXIT_SUCCESS;
}"
MAIN="./main.c"
touch MAIN
echo -n "$(eval echo $TXTMAIN)" >> "$MAIN"
gedit MAIN
This piece of code gives the following output when you edit main.c :
#include <stdlib.h> #include <stdio.h> int main(int argc, char const* argv[]) { printf("hello, world\n"); return EXIT_SUCCESS; }
However, when replacing line 13 by echo -n "$TXTMAIN_C" >> "$MAIN", it gives the right output :
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char const* argv[])
{
printf("hello, world\n");
return EXIT_SUCCESS;
}
I still don't know wether that's an echo or eval issue, or if there is a way around for my pointer-like problem.
Any advice is very welcome !

Answer 1

There are a few errors in your script, and it's more complicated than it should be.
If you want to use an indirect variable like that, use the ${!FOO} syntax, and put quotes where appropriate:
#!/bin/sh
#this is a short example in the case the user selected C as the language
TXTMAIN=TXTMAIN_C # don't force a $ here
TXTMAIN_C="#include <stdlib.h>
#include <stdio.h>
int main(int argc, char const* argv[])
{
printf(\"hello, world\n\");
return EXIT_SUCCESS;
}"
MAIN="./main.c"
echo "${!TXTMAIN}" > "$MAIN" # overwrite here, if you want to
# append, use >>. `touch` is useless