Short-circuiting overloaded && and || using expression templates

12:56:00 PM 0 Comments A+ a-

This blog post is just a quick note that C++ offers (at least) two distinct ways to represent lazy computation that is lexically in the same scope but may execute lazily at a later time. In doing so, the computation must capture the local context (i.e., variables) so that it can be used later when needed. Clearly, lambda expressions are a direct language supported mechanism for that. Closures that come out of a lambda expression often capture the context and of course some behavior to be run later. The second mechanism is about 20 years old (as of this writing): Expression Templates.

Lets take an example of short-circuiting overloaded && and || operators. Regular overloaded && and || do not short circuit in C++. The reason is that before calling the overloaded operator &&, both the left-hand-side and the right-hand-side arguments of the overloaded function are evaluated. A function call is a sequence-point and therefore all the computations and the side-effects are complete before making the function call. This is eager strategy.

Expression Templates is a library-only approach to defer computation at a later time while keeping the context of the original expression around. Sounds a lot like lambda expressions.

Consider the struct S below. I would like to implement short-circuiting && and || for this type.

struct S
{
  bool val;
  explicit S(bool b) : val(b) {}

  bool is_true () const 
  {
    return val;
  }
};

S operator && (const S & s1, const S & s2)
{
  return s1.is_true()? S{s1.val && s2.val} : s1;
}

int main(void)
{
  S s1{false}, s2{true}, s3{true};
  S s4 = s1 && s2 && s3; // false
}
There is hardly any optimization at all. The overloaded && operator is called twice no matter what. Although the result of the expression s1 && s2 && s3 is known just by looking at s1. An opportunity for optimization is wasted (if you ever wanted to optimize that way!).

So let's use expression templates. The trick is to convert the expression into a tree of recursively nested instantiations of the Expr template. The tree is evaluated separately after construction.

The following code implements short-circuited && and || operators for S as long as it provides logical_and and logical_or free functions and it is convertible to bool. The code is in C++14 but the idea is applicable in C++98 also.

#include <iostream>

struct S
{
  bool val;

  explicit S(int i) : val(i) {}  
  explicit S(bool b) : val(b) {}

  template <class Expr>
  S (const Expr & expr)
   : val(evaluate(expr).val)
  { }

  template <class Expr>
  S & operator = (const Expr & expr)
  {
    val = evaluate(expr).val;
    return *this;
  }

  bool is_true () const 
  {
    return val;
  }
};

S logical_and (const S & lhs, const S & rhs)
{
    std::cout << "&& ";
    return S{lhs.val && rhs.val};
}

S logical_or (const S & lhs, const S & rhs)
{
    std::cout << "|| ";
    return S{lhs.val || rhs.val};
}


const S & evaluate(const S &s) 
{
  return s;
}

template <class Expr>
S evaluate(const Expr & expr) 
{
  return expr.eval();
}

struct LazyAnd 
{
  template <class LExpr, class RExpr>
  auto operator ()(const LExpr & l, const RExpr & r) const
  {
    const auto & temp = evaluate(l);
    return temp.is_true()? logical_and(temp, evaluate(r)) : temp;
  }
};

struct LazyOr 
{
  template <class LExpr, class RExpr>
  auto operator ()(const LExpr & l, const RExpr & r) const
  {
    const auto & temp = evaluate(l);
    return temp.is_true()? temp : logical_or(temp, evaluate(r));
  }
};


template <class Op, class LExpr, class RExpr>
struct Expr
{
  Op op;
  const LExpr &lhs;
  const RExpr &rhs;

  Expr(const LExpr& l, const RExpr & r)
   : lhs(l),
     rhs(r)
  {}

  auto eval() const 
  {
    return op(lhs, rhs);
  }
};

template <class LExpr>
auto operator && (const LExpr & lhs, const S & rhs)
{
  return Expr<LazyAnd, LExpr, S> (lhs, rhs);
}

template <class LExpr, class Op, class L, class R>
auto operator && (const LExpr & lhs, const Expr<Op,L,R> & rhs)
{
  return Expr<LazyAnd, LExpr, Expr<Op,L,R>> (lhs, rhs);
}

template <class LExpr>
auto operator || (const LExpr & lhs, const S & rhs)
{
  return Expr<LazyOr, LExpr, S> (lhs, rhs);
}

template <class LExpr, class Op, class L, class R>
auto operator || (const LExpr & lhs, const Expr<Op,L,R> & rhs)
{
  return Expr<LazyOr, LExpr, Expr<Op,L,R>> (lhs, rhs);
}

std::ostream & operator << (std::ostream & o, const S & s)
{
  o << s.val;
  return o;
}

S and_result(S s1, S s2, S s3)
{
  return s1 && s2 && s3;
}

S or_result(S s1, S s2, S s3)
{
  return s1 || s2 || s3;
}

int main(void) 
{
  for(int i=0; i<= 1; ++i)
    for(int j=0; j<= 1; ++j)
      for(int k=0; k<= 1; ++k)
        std::cout << i << j << k << " " << and_result(S{i}, S{j}, S{k}) << std::endl;

  for(int i=0; i<= 1; ++i)
    for(int j=0; j<= 1; ++j)
      for(int k=0; k<= 1; ++k)
        std::cout << i << j << k << " " << or_result(S{i}, S{j}, S{k}) << std::endl;

  return 0;
}
Let's break it apart piece by piece.

Type S has new conversion and assignment operators that convert a generic Expr argument that is convertible to S. The expression is not evaluated until it is actually assigned to another S. We just call evaluate on the expression to begin execution of the computation wrapped inside Expr. logical_and and logical_or are free functions that perform the non-short-circuiting logical operations because we're going to hijack the overloaded && and || for short-circuiting.

The evaluate free functions take care of the trivial base case when Expr happens to just another S and all other cases when Expr is a compound expression.

struct LazyAnd and LazyOr are the short-circuiting && and ||. They always evaluate the left-hand-side but may not evaluate the right-hand-side if it is not required.

Expr template enables construction of so called expression templates. It is meant to be instantiated recursively. for example, an expression template for (s1 && s2) looks like Expr<LazyAnd, S, S> whereas for (s1 && s2 && s3) it is Expr<LazyAnd, Expr<LazyAnd, S , S>, S>. One last example: (s1 && (s2 && s3)) becomes Expr<LazyAnd, S, Expr<LazyAnd, S , S>>.
Of course, creating the nested Expr instantiations manually is berserk. So we use overloaded && and || operators that instead of computing the result eagerly, produce and expression that we can evaluate later. I've avoided writing overly generic && and || operator by using the second argument that is either S or and Expr. So the operator does not match with types outside those. Take a look at the examples above. It is fairly straightforward to see how an expressions turns into a tree. Note that construction of tree does not involve calling logical_and and logical_or functions

Finally, the assignment operator and copy-ctor of S take care of executing the expression. LazyAnd and LazyOr do the least possible work while ensuring that left-hand-side is always evaluated. Here is the output of the program. Checkout the live example here.
000 0
001 0
010 0
011 0
100 && 0
101 && 0
110 && && 0
111 && && 1
000 || || 0
001 || || 1
010 || 1
011 || 1
100 1
101 1
110 1
111 1
Bottom line: Expression templates and lambdas are both suitable for passing lazy computations to functions. They both can capture local context (variables) and don't extend the life-cycle of the captured argument. Their type is not meant to be observed (it is often unpronounceable). Expression templates, however, are very specific because they appear only in the context of overloaded operators and as a result they may be lot more expressive.

This blog post is motivate by this question on Stackoverflow. Also see comments on reddit/r/cpp.