String replace in wildcard using find command

Question

I want to exclude all C files that have a matching asm file. The assembly files all have _x86_64.S at the end. The C files have identical names, but with .c instead of _x86_64.S at the end. How would I do something like:

find . -not -name *{_x86_64.S but replace with .c}

So, for example, if my_func.c and my_func_x86_64.S exist, then the file my_func.c will be excluded by the above command.

Edit: The answer provided below works. I forgot to mention I was using this in a Makefile. What I had been doing was the following:

ASM_FILES = $(shell find ./src/assembly/x86_64/ -name "*.S" -printf "-not -name \"*%f*\" -and ")
ASM_INCLUDE += -wholename "./src/assembly/x86_64/*.S" -or
EXCLUDE += $(subst _x86_64.S,.c,$(ASM_FILES))

This works, but I was hoping for something a little cleaner.

Answer 1

Well, something like [ -e "${1%.c}.asm" ] would test if the .c file named by the first command line arg has a corresponding .asm.

So, remembering that find's -exec is a condition, this can be used to print all .c files without a corresponding .asm.

find . -name "*.c" -exec sh -c '! [ -e "${1%.c}.asm" ]' find-sh {} \; -print

Then, we'll have to include everything else (assuming you only want regular files), e.g.:

find . -type f \( ! -name "*.c" -o
                  -name "*.c" -exec sh -c '! [ -e "${1%.c}.asm" ]' find-sh {} \; \
               \) -print

e.g. given the four files bar.c foo.asm foo.c hello.txt, that prints

./hello.txt
./foo.asm
./bar.c

Though that may be slightly slow, since it forks a new shell instance for each .c file.

Alternatively, you could do that in a shell loop with ** for recursive globbing (needs shopt -s globstar in Bash):

$ shopt -s globstar
$ for f in ./**/* ; do
    case $f in *.c)
        if [ -e "${f%.c}.asm" ]; then continue; fi
    esac;
    printf -- "%s\n" "$f"
  done
./bar.c
./foo.asm
./hello.txt

(Note that there are differences between shells on how ** deals with symlinks within the tree.)