The use of $@ doesn't look right to me. In the trivial case exposed here where there are no additional arguments to pass to the .c program, the shell executes

The use of $@ doesn't look right to me. In the trivial case exposed here where there are no additional arguments to pass to the .c program, the shell executes gcc "print.c" -o .out && exec ./.out and it works "by chance". In a more complex scenario where print.c expects some parameters, it won't work. For example, ./print.c a b c will result in the shell trying to invoke gcc "print.c" -o "a" "b" "c".out && exec ./"a" "b" "c".out which makes no sense. Are you sure you didn't intend $0 instead of $@ ?