When I include another source(I.e stdio.h) the preprocessor is smart enough to include only the functions that I am using in my code?
Example: Assuming this small program, would be ease to include only what I am using, and what the printf functions uses, including them recursively, but what about bigger programs?
#include <stdio.h>
int main(void) {
printf("Hello World\n");
return 0;
}
No. On the contrary:
#include performs textual replacement: it opens the file and copies all1 of its contents into your main C file. In the process it executes all preprocessor instructions in the included file. Amongst other things, this means that it will recursively include all files that are #included in the header.
#include does not know and does not care which part of the included file you end up using.
1 As mentioned, preprocessor instructions are executed in the included file. This can modify what gets included. For example, assume the following header file header.h:
#ifndef HEADER_H
#define HEADER_H
#ifdef NDEBUG
# define LOG(...) ((void) 0)
#else
# define LOG(...) log_message(__FILE__, __LINE__, __VA_ARGS__)
inline void log_message(const char* filename, int line, ...) {
// Logging code omitted for brevity.
}
#endif
// other stuff
#endif
Now, if your main.c file looks as follows:
#define NDEBUG
#include "header.h"
int main(void) {
// …
LOG("hello");
}
… then, after preprocessing, your main.c file would looks something like this (I’m omitting some irrelevant stuff):
# 1 "main.c"
# 1 "./header.h" 1
# 13 "./header.h"
// other stuff
# 3 "main.c" 2
int main(void) {
// …
((void) 0);
}
… in other words, only the part of header.h that corresponds to #ifdef NDEBUG was included, not the part in the #else clause. If we had included header.h without defining NDEBUG, then the included header code would have contained the definition of log_message.
As others have said, #include will paste verbatim the entire file you are targeting. However you normally include headers, which tend to look like
extern int a (int b);
extern char * c (void);
static inline int d (int e, int f) {
...
}
extern void * g (void * h);
...
The code above occupies exactly zero memory (unless you start using one of the inline functions), since it is entirely composed of instructions for the compiler and nothing else.
