GCC: How does make command work ?

Question

in the following code func.c :

 int Myfunc1(int i, int z)
 {
 return i;
 }

 int main()
 {
 int ans;

 /* casting the function into an 'int (int)' function */
 ans = ((int(*)(int))(Myfunc1))(5);

 printf("ans: %dnn", ans);

 return 0;
 }

I tried to cast an int(int,int) function into an int(int) function an got the gcc warning and note:

 func.c:13:32: warning: function called through a non-compatible type [enabled by default]
 func.c:13:32: note: if this code is reached, the program will abort

and when trying to run I get:

 Illegal instruction (core dumped)

But if i compile this file with a .cpp ending with the gcc compiler it works OK.

Answer 1

Make utility is used to create the binary or libraries. It is very helpful and avoid to write gcc command manually.
Usually a big may have thousands of files and it is very tedious to write gcc command manually to build up the binary.

In Simple term, it is described as:
Target: Dependency1 Dependency2
action required to achieve target.

For example:
I have following files in same directory.
my_file.h - Header file which includes function prototype
my_file.c - Function defined in my_file.c file
main.c - Function called from here.
myBin- binary or executable

Target: Dependency1 Dependency2
myBin: main.o my_file.o
gcc -I . -c main.o my_file.o

Target: Dependency1
main.o: main.c my_file.h
gcc -I . -c main.c

my_file.o: my_file.c my_file.h
gcc -I . -c my_file.c