问题
I have an array of edges, which is defined as a C-style array of doubles, where every 4 doubles define an edge, like this:
double *p = ...;
printf("edge1: %lf %lf %lf %lf\n", p[0], p[1], p[2], p[3]);
printf("edge2: %lf %lf %lf %lf\n", p[4], p[5], p[6], p[7]);
So I want to use std::sort() to sort it by edge length. If it was a struct Edge { double x1, y1, x2, y2; }; Edge *p;, I would be good to go.
But in this case, the double array has a block size that is not expressed by the pointer type. qsort() allows you to explicitly specify the block size, but std::sort() infers the block-size by the pointer type.
For performance reasons (both memory-usage and CPU), let's say that it's undesirable to create new arrays, or transform the array somehow. For performance reasons again, let's say that we do want to use std::sort() instead of qsort().
Is it possible to call std::sort() without wasting a single CPU cycle on transforming the data?
Possible approach:
An obvious approach is to try to force-cast the pointer:
double *p = ...;
struct Edge { double arr[4]; };
Edge *p2 = reinterpret_cast<Edge*>(p);
std::sort(...);
But how do I make sure the data is aligned properly? Also, how do I make sure it will always be aligned properly on all platforms and architectures?
Or can I use a typedef double[4] Edge;?
回答1:
How about having a reordering vector? You initialize vector with 1..N/L, pass std::sort a comparator that compares elements i1*L..i1*L+L to i2*L..i2*L+L, and when your vector is properly sorted, reorder the C array according to new order.
In response to comment: yes things get complicated, but it may just be good complication! Take a look here.
回答2:
You can use a "stride iterator" for this. A "stride iterator" wraps another iterator and an integer step size. Here's a simple sketch:
template<typename Iter>
class stride_iterator
{
...
stride_iterator(Iter it, difference_type step = difference_type(1))
: it_(it), step_(step) {}
stride_iterator& operator++() {
std::advance(it_,step_);
return *this;
}
Iter base() const { return it_; }
difference_type step() const { return step_; }
...
private:
Iter it_;
difference_type step_;
};
Also, helper functions like these
template<typename Iter>
stride_iterator<Iter> make_stride_iter(
Iter it,
typename iterator_traits<Iter>::difference_type step)
{
return stride_iterator<Iter>(it,step);
}
template<typename Iter>
stride_iterator<Iter> make_stride_iter(
stride_iterator<Iter> it,
typename iterator_traits<Iter>::difference_type step)
{
return stride_iterator<Iter>(it.base(),it.step() * step);
}
should make it fairly easy to use stride iterators:
int array[N*L];
std::sort( make_stride_iter(array,L),
make_stride_iter(array,L)+N );
Implementing the iterator adapter all by yourself (with all operators) is probably not a good idea. As Matthieu pointed out, you can safe yourself a lot of typing if you make use of Boost's iterator adapter tools, for example.
Edit: I just realized that this doesn't do what you wanted since std::sort will only exchange the first element of each block. I don't think there's an easy and portable solution for this. The problem I see is that swapping "elements" (your blocks) cannot be (easily) customized when using std::sort. You could possibly write your iterator to return a special reference type with a special swap function but I'm not sure whether the C++ standard guarantees that std::sort will use a swap function that is looked up via ADL. Your implementation may restrict it to std::swap.
I guess the best answer is still: "Just use qsort".
回答3:
I don't remember exactly how to do this, but if you can fake anonymous functions, then you can make a comp(L) function that returns the version of comp for arrays of length L... that way L becomes a parameter, not a global, and you can use qsort. As others mentioned, except in the case where your array is already sorted, or backwards or something, qsort is going to be pretty much just as fast as any other algorithm. (there's a reason it's called quicksort after all...)
回答4:
It's not part of any ANSI, ISO, or POSIX standard, but some systems provide the qsort_r() function, which allows you to pass an extra context parameter to the comparison function. You can then do something like this:
int comp(void *thunk, const void *a, const void *b)
{
int L = (int)thunk;
// compare a and b as you would normally with a qsort comparison function
}
qsort_r(array, N, sizeof(int) * L, (void *)L, comp);
Alternatively, if you don't have qsort_r, you can use the callback(3) package from the ffcall library to create closures at runtime. Example:
#include <callback.h>
void comp_base(void *data, va_alist alist)
{
va_start_int(alist); // return type will be int
int L = (int)data;
const void *a = va_arg_ptr(alist, const void*);
const void *b = va_arg_ptr(alist, const void*);
// Now that we know L, compare
int return_value = comp(a, b, L);
va_return_int(alist, return_value); // return return_value
}
...
// In a function somewhere
typedef int (*compare_func)(const void*, const void*);
// Create some closures with different L values
compare_func comp1 = (compare_func)alloc_callback(&comp_base, (void *)L1);
compare_func comp2 = (compare_func)alloc_callback(&comp_base, (void *)L2);
...
// Use comp1 & comp2, e.g. as parameters to qsort
...
free_callback(comp1);
free_callback(comp2);
Note that the callback library is threadsafe, since all parameters are passed on the stack or in registers. The library takes care of allocating memory, making sure that memory is executable, and flushing the instruction cache if necessary to allow dynamically generated code (that is, the closure) to be executed at runtime. It supposedly works on a large variety of systems, but it's also quite possible that it won't work on yours, either due to bugs or lack of implementation.
Also note that this adds a little bit of overhead to the function call. Each call to comp_base() above has to unpack its arguments from the list passed it (which is in a highly platform-dependent format) and stuff its return value back in. Most of the time, this overhead is miniscule, but for a comparison function where the actual work performed is very small and which will get called many, many times during a call to qsort(), the overhead is very significant.
回答5:
For the new question, we need to pass in sort() a kind of iterator that will not only let us compare the right things (i.e. will make sure to take 4 steps through our double[] each time instead of 1) but also swap the right things (i.e. swap 4 doubles instead of one).
We can accomplish both by simply reinterpreting our double array as if it were an array of 4 doubles. Doing this:
typedef double Edge[4];
doesn't work, since you can't assign an array, and swap will need to. But doing this:
typedef std::array<double, 4> Edge;
or, if not C++11:
struct Edge {
double vals[4];
};
satisfies both requirements. Thus:
void sort(double* begin, double* end) {
typedef std::array<double, 4> Edge;
Edge* edge_begin = reinterpret_cast<Edge*>(begin);
Edge* edge_end = reinterpret_cast<Edge*>(end);
std::sort(edge_begin, edge_end, compare_edges);
}
bool compare_edges(const Edge& lhs, const Edge& rhs) {
// to be implemented
}
If you're concerned about alignment, can always just assert that there's no extra padding:
static_assert(sizeof(Edge) == 4 * sizeof(double), "uh oh");
回答6:
std::array< std::array<int, L>, N > array;
// or std::vector< std::vector<int> > if N*L is not a constant
std::sort( array.begin(), array.end() );
回答7:
I'm not sure if you can achieve the same result without a lot more work. std::sort() is made to sort sequences of elements defined by two random access iterators. Unfortunately, it determines the type of the element from the iterator. For example:
std::sort(&array[0], &array[N + L]);
will sort all of the elements of array. The problem is that it assumes that the subscripting, increment, decrement, and other indexing operators of the iterator step over elements of the sequence. I believe that the only way that you can sort slices of the array (I think that this is what you are after), is to write an iterator that indexes based on L. This is what sellibitze has done in the stride_iterator answer.
回答8:
namespace
{
struct NewCompare
{
bool operator()( const int a, const int b ) const
{
return a < b;
}
};
}
std::sort(array+start,array+start+L,NewCompare);
Do test with std::stable_sort() on realistic data-sets - for some data mixes its substantially faster!
On many compilers (GCC iirc) there's a nasty bite: the std::sort() template asserts that the comparator is correct by testing it TWICE, once reversed, to ensure the result is reversed! This will absolutely completely kill performance for moderate datasets in normal builds. The solution is something like this:
#ifdef NDEBUG
#define WAS_NDEBUG
#undef NDEBUG
#endif
#define NDEBUG
#include <algorithm>
#ifdef WAS_NDEBUG
#undef WAS_NDEBUG
#else
#undef NDEBUG
#endif
Adapted from this excellent blog entry: http://www.tilander.org/aurora/2007/12/comparing-stdsort-and-qsort.html
回答9:
Arkadiy has the right idea. You can sort in place if you create an array of pointers and sort that:
#define NN 7
#define LL 4
int array[NN*LL] = {
3, 5, 5, 5,
3, 6, 6, 6,
4, 4, 4, 4,
4, 3, 3, 3,
2, 2, 2, 2,
2, 0, 0, 0,
1, 1, 1, 1
};
struct IntPtrArrayComp {
int length;
IntPtrArrayComp(int len) : length(len) {}
bool operator()(int* const & a, int* const & b) {
for (int i = 0; i < length; ++i) {
if (a[i] < b[i]) return true;
else if (a[i] > b[i]) return false;
}
return false;
}
};
void sortArrayInPlace(int* array, int number, int length)
{
int** ptrs = new int*[number];
int** span = ptrs;
for (int* a = array; a < array+number*length; a+=length) {
*span++ = a;
}
std::sort(ptrs, ptrs+number, IntPtrArrayComp(length));
int* buf = new int[number];
for (int n = 0; n < number; ++n) {
int offset = (ptrs[n] - array)/length;
if (offset == n) continue;
// swap
int* a_n = array+n*length;
std::move(a_n, a_n+length, buf);
std::move(ptrs[n], ptrs[n]+length, a_n);
std::move(buf, buf+length, ptrs[n]);
// find what is pointing to a_n and point it
// to where the data was move to
int find = 0;
for (int i = n+1; i < number; ++i) {
if (ptrs[i] == a_n) {
find = i;
break;
}
}
ptrs[find] = ptrs[n];
}
delete[] buf;
delete[] ptrs;
}
int main()
{
for (int n = 0; n< NN; ++n) {
for (int l = 0; l < LL; ++l) {
std::cout << array[n*LL+l];
}
std::cout << std::endl;
}
std::cout << "----" << std::endl;
sortArrayInPlace(array, NN, LL);
for (int n = 0; n< NN; ++n) {
for (int l = 0; l < LL; ++l) {
std::cout << array[n*LL+l];
}
std::cout << std::endl;
}
return 0;
}
Output:
3555
3666
4444
4333
2222
2000
1111
----
1111
2000
2222
3555
3666
4333
4444
回答10:
A lot of these answers seem like overkill. If you really have to do it C++ style, using jmucchiello's example:
template <int Length>
struct Block
{
int n_[Length];
bool operator <(Block const &rhs) const
{
for (int i(0); i < Length; ++i)
{
if (n_[i] < rhs.n_[i])
return true;
else if (n_[i] > rhs.n_[i])
return false;
}
return false;
}
};
and then sort with:
sort((Block<4> *)&array[0], (Block<4> *)&array[NN]);
It doesn't have to be any more complicated.
来源：https://stackoverflow.com/questions/1649529/sorting-by-blocks-of-elements-with-stdsort