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laila [671]
3 years ago
12

The IBM system/370 architecture uses a two-level memory structure and refers to the two levels as segments and pages, although t

he segmentation approach lacks many of the features described in Chapter 8. For the basic 370 architecture, the page size may be either 2 KB or 4 KB, and the segment size is fixed at either 64 KB or 1 MB. For the 370/XA and 370/ESA architectures, the page size is 4 KB and the segment size is 1 MB. What advantages of segmentation does this scheme lack
Computers and Technology
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

Explanation:

Introductory thought:  

Thinking about the architecture for the IBM System/370 that is fit for giving higher preparing force and more stockpiling limit since it utilizes two-level of memory structure which is known as the segments and caches and the given plan is as per the following:  

  • In division the address space is isolated into quantities of segments of variable sizes and then again in paging the address space is partitioned into quantities of fixed-sized units that are known as pages.
  • It is more helpful for better utilization of memory.
  • The fundamental architecture for the IBM system/370 architecture the size of the page are for the most part either 2KiloBytes (KB) or 4KB and the size of the segments will be changed from 64KiloBytes to 1MegaBytes.  
  • The two dissimilar kinds of architecture that existed are 370/XA that is the Extended Architecture, also the other is 370/ESA known as the Enterprise Systems Architecture.  
  •  The page size is 4KB for the Extended variant and 1MB for the Enterprise Systems Architecture (ESA).

Advantages of segmentation that the given scheme needs to incorporate includes:

  • In the configuration of the architecture for IBM System/370 the segments are of a fixed size which isn't noticeable for the programmer and as such he will know the interior subtleties and subsequently inside architecture is stowed away from outside gentle.  
  • In IBM system/370 the reference bit is zero for the processor hardware as well as at whatever point another page is stacked into the edge, the reference bit is changed to one after referring to a specific area inside the casing.  
  • The quantity of lines of the page outline table is overseen by the Operating System and a page outline table section is moved to start with one line then onto the next depending on the length of the span of the referenced bit from the page outline which is sets to nothing.
  • Whenever there is a need to supplant the page then the page browsed the line of the longest life non referenced edges.
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Answer:

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The correct answer is E.

Explanation:

void change(int ar[], int low, inthigh)  {  

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  ar[high]= temp;  <em>// third, saves the element on temp in ar[high]. Complete switch.</em>

  change(ar,low + 1, high - 1);  <em>// Recursive call, adding one position to low, and subtracting one position to high. </em><em>Important: </em><em>When low and high have the same value, the recursive call will finish.</em>

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Answer:

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The find() method will find the position of day.

If we look at the string Happy holidays we see that the word day is at the 10th position of the string.

H     1

a      2

p      3

p      4

y      5

h      6

o      7

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