How to execute a certain part using makefile in C (LINUX) - c

Question

I am developing a C program that calls 2 functions, func1() and func2().
Here is the code
void func1(void){/*...*/}
void func2(void){/*...*/}
int main(){
func1();
func2();
return 0;
}
Is there a way that when I compile the program using make and a certain flag, it just executes func1() and if I compile using another flag then it only executes func2()?
How to make such a makefile? Can anyone provide me with a makefile that does this job?

Answer 1

You can use the -D flag of the compiler:
cc -o demo demo.c -Dfunc=func1
then
#include <stdio.h>
void func1(void) { puts("1"); }
void func2(void) { puts("2"); }
int main(void)
{
func();
return 0;
}
Output:
1
You can also pass the function as an argument to make:
CC = gcc
CFLAGS = -std=c11 -Wpedantic -Wall
action: demo.c
$(CC) $(CFLAGS) demo.c -o demo -Dfunc=$(argument)
clean:
rm -f demo
call make using:
make action argument=func1
or
make action argument=func2
But if you really want to do that:
int main(){
func1();
func2();
call both functions inside main, you should use another approach (although this solution seems quite sticky to me):
#include <stdio.h>
void func0(void) { /* do nothing */ }
void func1(void) { puts("1"); }
void func2(void) { puts("2"); }
#ifdef delete_func1
#define func1 func0
#endif
#ifdef delete_func2
#define func2 func0
#endif
int main(void)
{
func1();
func2();
return 0;
}
compile using:
cc -o demo demo.c -Ddelete_func1
or
cc -o demo demo.c -Ddelete_func2
then the Makefile should look like:
CC = gcc
CFLAGS = -std=c11 -Wpedantic -Wall
func1: demo.c
$(CC) $(CFLAGS) demo.c -o demo -Ddelete_func2
func2: demo.c
$(CC) $(CFLAGS) demo.c -o demo -Ddelete_func1
clean:
rm -f demo
call it using:
make func1
or
make func2
you can delete both functions using -D twice:
cc -o demo demo.c -Ddelete_func1 -Ddelete_func2