Pass Lines of Code as an Argument in C - c

Question

I'm working on a project in which I'm on a fairly strict word limit (I'm not 100% certain what qualifies as a word).
I'm looking to pass lines of code into a function as an argument, I've seen this done in JavaScript but I cannot find anything on it in C.
This is about what I'm looking for:
void onTime(int a, Code myCode) {
if(timer == a) {
//run myCode
}
}
And I could use it something like this:
onTime(45, {
//code
});
Is there a way in C that I can do something like this?

Answer 1

C doesn't allow inline code (referred to as lambda expressions in other languages) to be passed around. What you can do however is pass a pointer to an existing function.
// typedef for a function that takes no arguments and returns void
typedef void (*ftype)(void);
void func_to_run(void)
{
...
}
void onTime(int a, ftype code) {
if(timer == a) {
code();
}
}
You could then call it like this:
ontime(45, func_to_run);
If you want the function to be packaged with a set of values that it uses, similarly to C++ classes which have member functions which can access member variables, that involves some trickery.
Here's an example of how you might do something like this:
#include <stdio.h>
typedef int (*ftype)(int, int);
struct c {
int a;
int b;
ftype add_func;
ftype sub_func;
};
#define ADD(s) (s)->add_func((s)->a,(s)->b)
#define SUB(s) (s)->sub_func((s)->a,(s)->b)
int add(int a, int b)
{
return a+b;
}
int sub(int a, int b)
{
return a-b;
}
void run(struct c *op)
{
printf("add result=%d\n", ADD(op));
printf("sub result=%d\n", SUB(op));
}
int main()
{
struct c c1 = { 1, 2, add, sub };
struct c c2 = { 3, 4, add, sub };
struct c c3 = { 6, 1, add, sub };
struct c c4 = { 9, 4, add, sub };
run(&c1);
run(&c2);
run(&c3);
run(&c4);
}
Here we define a struct c which contains function pointers for two functions and two other variables.
The macros ADD and SUB are used to run the "member" functions of the struct and pass them the "private" members to use. It hides the fact that the actual function being called is independent of the parameters being passed to it.
This code outputs:
add result=3
sub result=-1
add result=7
sub result=-1
add result=7
sub result=5
add result=13
sub result=5
Of course, once you start doing this you're (crudely) doing what C++ is doing and would probably be better off switching to C++ which is made for OOP and now supports lambdas.