dir1 has dir2, file1.c and file1.h.
dir2 has file2.c
Now, if I want to access a function defined in file1.c in file2.c, I need to declare it in file1.h and inclu
including the .h files is not enough because it only includes the prototype of that function not the definition of the function and the definition of the function is in a seperate .c file.
one way to fix it is just type:
gcc -o out file1.c file2.c
or as Nate says you could seperate the compilation process and the link process
Now, if I want to access a function defined (in) file1.c in file2.c
A function defined in FILE1.c Access from FILE2.c example function: void sync(all){start sync ...}
Need on File2.h
AND Need on File2.c
Nothing else!
I always use also the keyword EXTERN in File2.h
Example:
The voted also work, choose with what you feel better, I feel better when I saw what happend during coding. Imagine an other team member has to review your code, it will be harder...
Compiling a c program happens in two steps basic steps: compiling and linking. Compiling turns source code into object code, and linking puts object code together, and ties all of the symbols together.
Your problem is a linker problem, not a compiler problem.
You are likely running the following:
gcc dir_2/file2.c
instead, do the following:
gcc -c dir_2/file2.c
gcc -c file1.c
gcc -o out file1.o file2.o
The reason this happens isn't because you didn't declare the function in the header, or didn't include the header properly. When the linker tries to put all the symbols together in the executable, it can't find your function because you are only linking half of your program.