Cannot execute output files( able to execute a.out but not when I make an output file with "make")

Question

Question has been solved , visit the bottom part of this paragraph for details.

I have a program: foo.cpp and bar.cpp and have created a Makefile to compile them.

This is how it(Makefile) looks:

CC=g++
CFLAGS=-c

foo.o: foo.cpp
    $(CC) $(CFLAGS) foo.cpp -o foo 
    
all: foo.o bar.o 

bar.o: bar.cpp
    $(CC) $(CFLAGS) bar.cpp -o bar

Now, when I run make all, it compiles and I have the output files there but when I run ./foo it gives me this error--> bash: ./foo: Permission denied

However if I do g++ foo.cpp and then do ./a.out then it runs.

I have seen the similar questions like: https://askubuntu.com/questions/466605/cannot-open-output-file-permission-denied and tried the solution but it did not work.

I don't understand why this is happening. Can someone please help(I am new to Makefiles)?

Thanks

For Future visitors

Solution and Mistakes done:

1)using -c flag , it created an output file and not an executable file which I was trying to execute.

2)for more information , here is an answer on linkers: What's an object file in C?

Answer 1

The file foo is the object file generated from foo.cpp . Object files can of course not be executed, since they are not complete programs.

Since you don't give a specific name for the executable, the default is a.out on POSIX environments (like Linux).

If you want a specific file-name then you need to add a command to create it. And make sure it doesn't clash with an existing object file.

---

I recommend you change your whole Makefile to something much more simple, like this:

all: foo

foo: foo.o bar.o
    $(CXX) foo.o bar.o -o foo

That's all that is needed. The object files will be created, with the correct names , from implicit rules.