I am playing around with fold expression in c++17 with the latest clang++. I tried to implement the less operator for an array using this which I want to use for fixed size strings.
Here is where I got to. Is there a better way to do this, especially avoiding assigning the index in the expression?
Compiling this using "clang++ test_fold_expr_less.cpp -o test_fold_expr_less -std=c++1z" and the output is here.
prompt$ ./test_fold_expr_less
=== less ===
0
1
0
0
1
0
0
0
0
1
1
1
#include <iostream>
#include <utility>
std::uint64_t arr1[8] = {1, 7, 2, 4, 8, 9, 3, 6};
std::uint64_t arr2[8] = {1, 7, 2, 4, 8, 9, 3, 6};
std::uint64_t arr3[8] = {1, 7, 2, 5, 8, 9, 3, 6};
std::uint64_t arr4[8] = {1, 7, 2, 3, 8, 9, 3, 6};
struct less_t
{
template < typename T, std::size_t N, std::size_t... I >
bool impl(T const (& lhs)[N], T const (& rhs)[N], std::index_sequence < I... >) const
{
std::size_t i{0};
if (((i = I, (lhs[I] < rhs[I]) ? true : lhs[I] != rhs[I]) || ...))
return lhs[i] < rhs[i];
else
return false;
}
template < typename T, std::size_t N >
bool operator () (T const (& lhs)[N], T const (& rhs)[N]) const
{
return impl(lhs, rhs, std::make_index_sequence < N >());
}
};
int main()
{
std::cout << "=== less ===" << std::endl;
less_t const less{};
std::cout << less(arr1, arr2) << std::endl;
std::cout << less(arr1, arr3) << std::endl;
std::cout << less(arr1, arr4) << std::endl;
std::cout << less(arr2, arr1) << std::endl;
std::cout << less(arr2, arr3) << std::endl;
std::cout << less(arr2, arr4) << std::endl;
std::cout << less(arr3, arr1) << std::endl;
std::cout << less(arr3, arr2) << std::endl;
std::cout << less(arr3, arr4) << std::endl;
std::cout << less(arr4, arr1) << std::endl;
std::cout << less(arr4, arr2) << std::endl;
std::cout << less(arr4, arr3) << std::endl;
}
I'll start with some observations and hypotheses concerning the use of fold-expressions here.
Essentially, we want to write a lexicographical comparison via a single fold-expression. We want to bail out of the comparisons once we can determine the result. Using a unary left fold
(... OP comparison)
evaluates to
( ( (comparison(0) OP comparison(1)) OP comparison(2) )... )
The only ways to not evaluate the comparison(I)
are throwing an exception in one of the comparisons executed earlier and short-circuiting. I don't think using exceptions is a good idea here. So I'll try short-circuiting. This requires OP
to be either ||
or &&
, and the expression to the left must evaluate to a bool that tells the computation whether or not to continue:
( ( (comparison(0) && comparison(1)) && comparison(2) )... )
comparison(N) -> bool // continue?
In lexicographical comparison, we continue evaluating if the current elements of the lhs and rhs are equal. So the value of comparison(I)
must be lhs[I] == rhs[I]
:
#define comparison(I) (lhs[I] == rhs[I])
The result of the whole fold-expression then tells us whether or not both sequences are entirely equal:
auto equal = (... && comparison(I));
However, by doing this, we lose the information about whether lhs[I] < rhs[I]
or lhs[I] > rhs[I]
was the reason we stopped the comparison. We can get this information out of the expression by using a side-effect:
#define comparison(I) (less = lhs[I] < rhs[I], lhs[I] == rhs[I])
bool less;
auto equal = (... && comparison(I));
At this point, we know either equal == true
or we can use the value last stored to less
to determine the overall result:
return !equal && less;
(If lhs == rhs
then !(lhs < rhs)
, so we return false. Otherwise, lhs != rhs
and we use the result stored in less
. Therefore, less
doesn't need to be initialized if we have at least one element in the array.)
There is still room for improvement: We can perform the assignment to less
, and with it the value computation of lhs[I] < rhs[I]
only when we need it, that is, only when lhs[I] != rhs[I]
. Using another short-circuit:
#define comparison(I) (lhs[I] == rhs[I] || (less = lhs[I] < rhs[I], false))
// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I'd consider this quite cryptic. The value of the underlined part is always false
due to the use of a comma-expression. Since false
is the neutral element of ||
, the whole || (..)
only performs a side-effect but doesn't change the value of comparison(I)
, but this side-effect is only performed if the left-hand side of ||
yields false
.
One last improvement can be achieved by recognizing that if we never assign to less
, we know that lhs == rhs
and we can return false
.
Combined, we get:
bool less = false;
auto equal = (... && (lhs[I] == rhs[I] || (less = lhs[I] < rhs[I], false)));
(void)equal; // we don't need you
return less;
Complete example:
#include <iostream>
#include <utility>
struct loud
{
std::uint64_t v;
loud(int x) : v(x) {}
static auto& opEqual() { static int v = 0; return v; }
static auto& opLess() { static int v = 0; return v; }
friend bool operator==(loud l, loud r) { return opEqual()++, l.v == r.v; }
friend bool operator<(loud l, loud r) { return opLess()++, l.v < r.v; }
static void print_stats(std::ostream& o) {
o << "operator< " << opLess() << " operator== " << opEqual();
}
static void reset_stats() {
opEqual() = opLess() = 0;
}
};
loud arrs[][8] = {
{1, 7, 2, 4, 8, 9, 3, 6}
,{1, 7, 2, 4, 8, 9, 3, 6}
,{1, 7, 2, 5, 8, 9, 3, 6}
,{1, 7, 2, 3, 8, 9, 3, 6}
};
struct less_t
{
template < typename T, std::size_t N, std::size_t... I >
bool impl(T const (& lhs)[N], T const (& rhs)[N], std::index_sequence < I... >) const
{
bool less = false;
auto equal = (... && (lhs[I] == rhs[I] || (less = lhs[I] < rhs[I], false)));
(void)equal; // we don't need you
return less;
}
template < typename T, std::size_t N >
bool operator () (T const (& lhs)[N], T const (& rhs)[N]) const
{
return impl(lhs, rhs, std::make_index_sequence < N >());
}
};
template<class T, int N>
void test(T const (&lhs)[N], T const (&rhs)[N])
{
auto const stdres = std::lexicographical_compare(lhs, lhs+N, rhs, rhs+N);
loud::reset_stats();
auto const foldres = less_t{}(lhs, rhs);
std::cout << (stdres == foldres) << " -- ";
loud::print_stats(std::cout);
std::cout << "\n";
}
int main()
{
std::cout << std::boolalpha;
std::cout << "=== less ===" << std::endl;
for(auto& lhs : arrs)
for(auto& rhs : arrs)
test(lhs, rhs);
}
Output -- note I do not print the result of the fold-expression-function, but I compare that result to the result of std::lexicographical_compare
. true
therefore means both algorithms produced the same result.
=== less ===
true -- operator< 0 operator== 8
true -- operator< 0 operator== 8
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 0 operator== 8
true -- operator< 0 operator== 8
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 0 operator== 8
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 1 operator== 4
true -- operator< 0 operator== 8
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