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navik [9.2K]
3 years ago
5

What type of data would a RDBMS store in a linked list, two-way linked list, or B-trees? What are the advantages and disadvantag

e of these data structures?
Computers and Technology
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer:

* Two-way linked list is stored in a linked list.

Advantages of a link list

* linked list is a variety of linked list wherein route is conceivable both ways, either forward and in reverse effectively.

* They are the simplest data structure

Disadvantages of a link list

*There are difficulties in reverse traversing

*They use more memory than array

Explanation:

linked list is a straight assortment of information components whose request isn't given by their actual situation in memory. All things being equal, every component focuses to the following. It is a data structure comprising of an assortment of hubs which together speak to a grouping. In its most fundamental structure, every hub contains: data, and a reference (all in all, a connection) to the following hub in the succession.

linked list is an arrangement of connections which contains things. Each connection contains an association with another connection. linked list is the second most used data structure after exhibit.

This structure takes into consideration productive inclusion or expulsion of components from any situation in the arrangement during emphasis. More perplexing variations add extra connections, permitting more productive inclusion or evacuation of hubs at subjective positions. A downside of connected records is that entrance time is direct (and hard to pipeline). Quicker access, for example, arbitrary access, isn't attainable.

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E-Commerce edge

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3 years ago
Suppose the size of process is 10000 bytes and the relocation register is loaded with value 5000 which of the following memory a
natulia [17]

Answer:

The page field is 8-bit wide, then the page size is 256 bytes.

Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.

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3 years ago
What year did buck tooth bob become famous
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Answer:

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Explanation:

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