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1 year ago

What dictionary operation can you use to remove all the keys within the dictionary-named contacts?

1 answer:

3 0

The dictionary operation that you use to remove all the keys within the dictionary-named contacts is contacts.clear().

<h3>What is dictionary add and remove element from dictionary?</h3>

The act of Removing elements from Dictionary is one that can delete a specific item in a dictionary by the use of the pop() method.

Note that this method deletes an item with the given key and returns the value .

Hence, The dictionary operation that you use to remove all the keys within the dictionary-named contacts is contacts.clear().

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Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.

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The number of bits in the physical address is 26 bits. The number of entries in a page table is $\mathbf{2^{20}}$ entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

32-bit logical address
Page size = 4 KB = $\mathbf{2^{12} \ bytes }$
Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = $\mathbf{2^{26} \ bytes}$

Thus, the number of bits in the physical address is computed as:

= $\mathbf{log_2 \{Physical-memory-size\}}$

$=\mathbf{log_2(2^{26})}}$

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The number of entries in a page table = logical address space size/page size

The number of entries in a page table $\mathbf{= \dfrac{2^{32}}{2^{12}}}$

$\mathbf{=2^{20}}$ entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= $\mathbf{2^{20}\times 4 \ bytes}$

= 4 MB

suppose that this system supports up to 2^30 bytes of physical memory.

The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is $\mathbf{2^{20}}$ entries, and the size of the page table in a one-level paging scheme is 4MB.

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