Continuations/Coroutines/Generators in C++ / gcc / linux - c++

Question

Background: I'm trying to figure out how to implement continuations/coroutines/generators (whatever the following is called) by posing this toy problem. The environment is C++11 on gcc 4.6 and linux 3.0 x86_64. Non-portable is fine but using an external library (boost.coroutine, COROUTINE, etc) is not allowed. I think longjmp(3) and/or makecontext(2) and friends may help but not sure.
Description:
The following toy parser is supposed to parse sequences of as and bs of equal length. ie
((a+)(b+))+
such that the length of the second bracketed production equals the third.
When it finds a production (eg aaabbb) it outputs the number of as it finds (eg 3).
Code:
#include <stdlib.h>
#include <iostream>
using namespace std;
const char* s;
void yield()
{
// TODO: no data, return from produce
abort();
}
void advance()
{
s++;
if (*s == 0)
yield();
}
void consume()
{
while (true)
{
int i = 0;
while (*s == 'a')
{
i++;
advance();
}
cout << i << " ";
while (i-- > 0)
{
if (*s != 'b')
abort();
advance();
}
}
}
void produce(const char* s_)
{
s = s_;
// TODO: data available, continue into consume()
consume();
}
int main()
{
produce("aaab");
produce("bba");
produce("baa");
produce("aabbb");
produce("b");
// should print: 3 1 4
return 0;
}
Problem:
As you can see the state of the consume call stack must be saved when yield is called and then produce returns. When produce is called again, consume must be restarted by returning from yield. The challenge would be to modify the way produce calls consume, and implement yield so they function as intended.
(Obviously reimplementing consume so that it saves and rebuilds its state defeats the purpose of the exercise.)
I think what needs to be done is something like the example on the bottom of the makecontext man page: http://www.kernel.org/doc/man-pages/online/pages/man3/makecontext.3.html, but its not clear how to translate it onto this problem. (and I need sleep)
Solution:
(Thanks to Chris Dodd for design)
#include <stdlib.h>
#include <iostream>
#include <ucontext.h>
using namespace std;
const char* s;
ucontext_t main_context, consume_context;
void yield()
{
swapcontext(&consume_context, &main_context);
}
void advance()
{
s++;
if (*s == 0)
yield();
}
void consume()
{
while (true)
{
int i = 0;
while (*s == 'a')
{
i++;
advance();
}
cout << i << " ";
while (i-- > 0)
{
advance();
}
}
}
void produce(const char* s_)
{
s = s_;
swapcontext(&main_context, &consume_context);
}
int main()
{
char consume_stack[4096];
getcontext(&consume_context);
consume_context.uc_stack.ss_sp = consume_stack;
consume_context.uc_stack.ss_size = sizeof(consume_stack);
makecontext(&consume_context, consume, 0);
produce("aaab");
produce("bba");
produce("baa");
produce("aabbb");
produce("b");
// should print: 3 1 4
return 0;
}

Answer 1

Its fairly straight-forward to use makecontext/swapcontext for this -- you use makecontext to create a new coroutine context and swapcontext to swap between them. In you case, you need one additional coroutine to run the consume infinite loop, and you run main and produce in the main context.
So main should call getcontext+makecontext to create a new context that will run the consume loop:
getcontext(&consume_ctxt);
// set up stack in consume_context
makecontext(&consume_ctxt, consume, 0);
and then produce will switch to it instead of calling consume directly:
void produce(const char* s_)
{
s = s_;
swapcontext(&main_ctxt, &consume_ctxt);
}
and finally yield just calls swapcontext(&consume_ctxt, &main_ctxt); to switch back to the main context (which will continue in produce and immediately return).
Note that since consume is an infinite loop, you don't need to worry too much about what happens when it returns (so the link will never be used)