Continuations and Continuation Passing¶
1. Tail recursive functions¶
Consider the two functions below. Each one takes in a list of non-negative integers and uses recursion to return the product of the numbers in the list.
var product1 = function (ns) {
if (fp.isNull(ns)) {
return 1;
} else {
return fp.mul(fp.hd(ns),
product1(fp.tl(ns)));
}
};
var product2 = function (ns) {
var helper = function (ns,a) {
if (fp.isNull(ns)) {
return a;
} else {
return helper(fp.tl(ns),
fp.mul(a,fp.hd(ns)));
}
};
return helper(ns,1);
};
Which version is better and why? To begin answering this question, we need the following definition. A tail call is a function call performed as the final action in a function’s body. A function is tail recursive if all of the recursive calls it makes are tail calls. We alluded to the notion of tail recursion in the section on the reduce function. Now we will learn more about it and its advantages.
The problem below is about a recursive function to compute the greatest common divisor (or \(gcd\)) of two integers.
2. Continuation-Passing Style¶
Tail-call elimination or optimization (TCO) automatically removes tail calls. Two questions immediately arise. Why would we want to do this? Assuming we understand why, how could it be done?
To understand “why”, we first note that the product-of-numbers-in-a-list computation should return 0 with no computation if the input list contains one or more zeros.
With the product1 function, no computations are performed until the base case is reached. But what if the last number is the first zero in the input list? We could save the last recursive call, but we still need to return 0 all the way back, through the (possibly very high) call stack, doing an unnecessary multiplication by 0 upon each return.
With the product2 function, TCO removes the sequence of recursive calls and thus allows us to return right away when the first zero is found. However, since the computations are performed before each recursive call, a lot of effort may have been wasted before the first zero is detected.
In such cases, we would like to have better control over the flow of execution of our programs. For example, we would like to return immediately as soon as a zero is detected, as in TCO’ed product2, but, as in product1, we do not want to perform any expensive computations until we know for sure that no zeros are present in the input list.
Continuations allow us to make the control flow and execution state of our programs explicit and thus solve this problem.
A continuation is a callback function \(k\) that represents the current state of the program’s execution. More precisely, the continuation \(k\) is a function of one argument, namely the value that has been computed so far, that returns the final value of the computation after the rest of the program has run to completion.
So, at any point in the computation for our product example, the continuation \(k\) takes as input the partial product \(x\) that has been calculated so far and returns the final product. Of course, the final product is still computed recursively. So \(k\) only computes one multiplication and then calls another continuation to process the rest of the list.
For example, if the input list is [2,3,4,5,6], the continuation after the 3 has been processed is the function that takes \(x=6=2\times 3\) as input, computes the next product (namely, \(x \times 4\)) and finally passes this new value as input to the continuation for the next recursive call. [Note: In reality, in this case, the product will be computed in reverse, from the end of the list toward its head, but that is a detail at this point].
To see how the notion of a continuation actually leads to a new tail-recursive version of the product function, read through the description of the product3 function in the following slide show.
The next slide show offers a recap of the three versions of the product function that we’ve seen so far.
In addition to providing a technique that guarantees TCO can be performed, CPS offers a couple of other advantages over straightforward recursion and the accumulation technique. First, suppose we want to make sure that no unnecessary computations are performed when the input list contains a zero. We can define a new and improved version of the function, which is called product4 and appears below.
var product4 = function (ns) {
var cps_zero = function (ns,k) {
if (fp.isNull(ns)) {
return k(1);
} else if (fp.isEq(fp.hd(ns),0)) {
return 0; // *** the continuation is never invoked! ***
} else {
return cps_zero(fp.tl(ns),
function (x) {
return k(fp.mul(x,fp.hd(ns)));
});
}
};
return cps_zero(ns, function (x) { return x; });
};
Note that, although we could add a similar case to return 0 in the product1 function, the 0 that we return would be unnecessarily used in computations multiple times as we unwind from recursion. We could also add a similar “return 0” case in product2, but potentially many unnecessary multiplications would have already been performed on the accumulator argument by the time that zero was encountered.
To illustrate one more neat aspect of functions that use continuation-passing style, recall that negative numbers are not allowed in the input list. Hence we could view the erroneous appearance of a negative number in the list as an exception, for which we would want to immediately throw an error message and abandon the computation of the product without doing any multiplications. Using continuation-passing style to handle exceptions in this fashion is illustrated in the product5 version of the function below.
var product5 = function (ns) {
var cps_exception = function (ns,k) {
if (fp.isNull(ns)) {
return k(1);
} else if (fp.isEq(fp.hd(ns),0)) {
return 0;
} else if (fp.isLT(fp.hd(ns),0)) {
return "Negative numbers are not allowed.";
} else {
return cps_exception(fp.tl(ns),
function (x) {
return k(fp.mul(x,fp.hd(ns)));
});
}
};
return cps_exception(ns, function (x) { return x; });
};
Adding such an exception-handling case that returns a string would be impossible in product1 since that string would have to participate in all the multiplications that occur as we unwind from recursion. Although we could add such a case in product2, it would defeat one of the main goals of exception handling, namely to protect the values of critical variables from “damage” that may have occurred before the exception was encountered. Although product2 is simple enough as to not have any damaging side effects that could occur prior to an exception, in more complicated situations the accumulator technique could not avoid this because it performs computations as we descend into recursive calls. In contrast, product5 has performed absolutely no computations when the exception is encountered. Instead all it has done is to have partially defined the continuation function, which we can harmlessly decide to not call upon encountering the exception.
3. Continuation-Passing Style Practice Problem (Part 1)¶
The following problem is the first one in a sequence of three problems that require you to complete the implementation of a recursive function that uses continuation-passing style programming. This problem uses the \(gcd\) function introduced in the first problem in this set, but you do not need to remember how it was implemented.
4. Continuation-Passing Style Practice Problem (Part 2)¶
The following problem is the second one in a sequence of three problems that require you to complete the implementation of a recursive function that uses continuation-passing style programming. This problem uses the \(gcd\) function introduced in the first problem in this set, but you do not need to remember how it was implemented.
5. Continuation-Passing Style Practice Problem (Part 3)¶
The following problem is the last one in a sequence of three problems that require you to complete the implementation of a recursive function that uses continuation-passing style programming. This problem uses the \(gcd\) function introduced in the first problem in this set, but you do not need to remember how it was implemented.
6. More CPS Practice¶
This randomized review problem will give you more practice writing recursive functions in the continuation-passing style. To get credit for it, you must solve it correctly three times in a row.
