How would one implement Lazy Evaluation in C?

Question

Take for example,

The follow python code:

def multiples_of_2():
  i = 0
  while True:
    i = i + 2
    yield i

How do we translate thi

Answer 1

I have implemented my own lazy eval, with respects to solving the hamming's problem.

Heres my code for anyone whos interested:

#include <stdio.h>
#include <stdlib.h>

// Hamming problem in C.

typedef struct gen {
  int tracker;
  int (*next)(struct gen *g);
} generator;

int twos_gen(struct gen *g) {
  g->tracker = g->tracker + 2;
  return g->tracker;
}

generator* twos_stream() {
  generator *g = malloc(sizeof(generator));
  g->next = twos_gen;
  g->tracker = 0;
  return g;
}

int threes_gen(struct gen *g) {
  g->tracker = g->tracker + 3;
  return g->tracker;
}

generator* threes_stream() {
  generator *g = malloc(sizeof(generator));
  g->next = threes_gen;
  g->tracker = 0;
  return g;
}

int fives_gen(struct gen *g) {
  g->tracker = g->tracker + 5;
  return g->tracker;
}

generator* fives_stream() {
  generator *g = malloc(sizeof(generator));
  g->next = fives_gen;
  g->tracker = 0;
  return g;
}

int smallest(int a, int b, int c) {
  if (a < b) {
    if (c < a) return c;
    return a;
  }
  else {
    if (c < b) return c;
    return b;
  }
}

int hamming_gen(struct gen *g) {
  generator* twos = twos_stream();
  generator* threes = threes_stream();
  generator* fives = fives_stream();

  int c2 = twos->next(twos);
  int c3 = threes->next(threes);
  int c5 = fives->next(fives);

  while (c2 <= g->tracker) c2 = twos->next(twos);
  while (c3 <= g->tracker) c3 = threes->next(threes);
  while (c5 <= g->tracker) c5 = fives->next(fives);

  g->tracker = smallest(c2,c3,c5);
  return g->tracker;
}

generator* hammings_stream() {
  generator *g = malloc(sizeof(generator));
  g->next = hamming_gen;
  g->tracker = 0;
  return g;
}

int main() {
  generator* hammings = hammings_stream();
  int i = 0;
  while (i<10) {
    printf("Hamming No: %d\n",hammings->next(hammings));
    i++;
  }
}

Answer 2

You could try to encapsulate this in a struct:

typedef struct s_generator {
    int current;
    int (*func)(int);
} generator;

int next(generator* gen) {
    int result = gen->current;
    gen->current = (gen->func)(gen->current);
    return result;
}

Then you define you multiples with:

int next_multiple(int current) { return 2 + current; }
generator multiples_of_2 = {0, next_multiple};

You get the next multiple by calling

next(&multiples_of_2);

Answer 3

int multiples_of_2() {
    static int i = 0;
    i += 2;
    return i;
}

The static int i behaves like a global variable but is visible only within the contect of multiples_of_2().