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As it was implemented, you compare collections using IStructuralComparable, which requires you to pass in an IComparer to use in comparing individual elements from the two structures. Since IComparer and IStructuralComparable are two different interfaces, you have problems if the elements of the IStructuralComparable are themselves collections, which now get compared with IComparer.
But if you prefered IStructuralComparable to IComparer at the top level, why don't you prefer it now?
The IStructuralComparable interface supports only structural comparisons for sorting or ordering. The IStructuralEquatable interface supports custom comparisons for structural equality.
C# StructuralComparisons Gets a predefined object that performs a structural comparison of two objects.
As far as I see this is only exposed through the StructuralComparisons class. The only way I dirilik figure out to make this useful is to make a StructuralEqualityComparer C# StructuralComparisons Kullanımı helper class birli follow:
Being able to specify IStructuralEquatable/IStructuralComparable in such cases is actually useful. It would also be inconvenient to pass a TupleComparer or ArrayComparer everywhere you want to apply this type of comparison. The two approaches are not mutually exclusive.
Yöntemin bu uygulaması, bir derme nesnesinin Compare her öğesini diğer bir koleksiyon nesnesinin ait öğesiyle kontralaştırdığında zirdaki kabil davranır:
GetHashCode Metodu: Nesnenin yapısal özelliklerine dayalı olarak bir hash kodu üretir. Bu, koleksiyonların hızlı bir şekilde hakkındalaştırılmasını sağlamlar.
My understanding is that it's used for collection C# StructuralComparisons Kullanımı like types, and encapsulates the structural part of the comparison, but leaves the comparison of the elements to a comparer passed in by the user. (link) (Got from the comments section of the question)
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For example, An implementation of IStructureComparable for a list would only C# StructuralComparisons Nedir need to scan a flat list whereas an implementation for a binary tree would also need to take into account the branching of the tree (e.