Here follows a community-wiki copy of the self-answer, which can be chosen as "the" answer. Please direct up/downvotes to the actual self-answer
Here is the method I found:
/* Example calculates the sum of the prime factors of the first 32 Fibonacci numbers */
#include <stdio.h>
typedef enum{false=0, true=1}bool;
/* the following line is the only time I have ever required "auto" */
#define FOR(i,iterator) auto bool lambda(i); yield_init = (void *)λ iterator; bool lambda(i)
#define DO {
#define YIELD(x) if(!yield(x))return
#define BREAK return false
#define CONTINUE return true
#define OD CONTINUE; }
/* Warning: _Most_ FOR(,){ } loops _must_ have a CONTINUE as the last statement.
* * Otherwise the lambda will return random value from stack, and may terminate early */
typedef void iterator; /* hint at procedure purpose */
static volatile void *yield_init;
#define YIELDS(type) bool (*yield)(type) = yield_init
iterator fibonacci(int n){
YIELDS(int);
int i;
int pair[2] = {0,1};
YIELD(0); YIELD(1);
for(i=2; i<n; i++){
pair[i%2] = pair[0] + pair[1];
YIELD(pair[i%2]);
}
}
iterator factors(int n){
YIELDS(int);
int i;
for(i=2; i*i<=n; i++){
while(n%i == 0 ){
YIELD(i);
n/=i;
}
}
YIELD(n);
}
main(){
FOR(int i, fibonacci(32)){
printf("%d:", i);
int sum = 0;
FOR(int factor, factors(i)){
sum += factor;
printf(" %d",factor);
CONTINUE;
}
printf(" - sum of factors: %d\n", sum);
CONTINUE;
}
}
Got the idea from http://rosettacode.org/wiki/Prime_decomposition#ALGOL_68 - but it reads better in C