OrderedIndex.opBinaryRight

in operator. Check to see if the given element exists in the container.

Complexity: O(log(n))

  1. bool opBinaryRight(Elem e)
  2. bool opBinaryRight(K k)
    mixintemplate OrderedIndex(size_t N, bool allowDuplicates, alias KeyFromValue, alias Compare, ThisContainer)
    static if(!isImplicitlyConvertible!(KeyType, Elem))
    bool
    opBinaryRight
    (
    string op
    K
    )
    (
    K k
    )
    if (
    op == "in" &&
    isImplicitlyConvertible!(K, KeyType)
    )

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