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Lelu [443]
2 years ago
6

State one technique for overcoming external fragmentation in dynamic partitioning. Why is this technique of overcoming external

fragmentation considered inefficient
Computers and Technology
1 answer:
prisoha [69]2 years ago
4 0

Compaction often gives solutions to the issues regarding external fragmentation.

One technique for overcoming external fragmentation in dynamic partitioning is

Compaction.

The reason why this technique of overcoming external fragmentation may be inefficient is because:

  • External fragmentation may need a lot of compaction and it is an expensive operation.

  • The use of contiguous allocation is often hard to fit processes into memory and also it is so difficult to grow or shrink the amount of memory allocated to a process.

  • Compaction only takes place when relocation is dynamic, and this also is expensive.

Compaction often shuffle memory notes or contents and then put or pile them up all in free memory and in one large block.

External fragmentation takes place when free memory is removed into small blocks.

Learn more from

brainly.com/question/23636593

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Answer:

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endf_input = '1'

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   student_file[name] = st_id

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if search == 0:

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for key, value in student_file.items():

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Explanation:

The program source code uses a for loop to continuously add data to the student file dictionary structure but breaks from the loop when a zero is received as input. The program enters search mode and returns the name of the student with the searched id number.

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2 years ago
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igor_vitrenko [27]

Answer:

#include <iostream>

# include <conio.h>

using namespace std;

main()

{

int size;

cout<<"enter the size of array";

cin>>size;

int a[size],b[size];

for (int i=0;i<size;i++)

{

cout<<"enter the value in array a"<<i;

cin>>a[i];

}

for (int k=0;k<=size;k++)

{

{

 if (a[k]%2==0)

 {

  for (int l=k; l<=size;l++)

  {

   

  a[l]=a[l+1];

   

   

     }

     size=size-1;

 }

   

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cout<<"\nArray list without even";

for(int j=0;j<size+1;j++)

{

cout<<"\n"<<a[j];

}

getch();

}

Explanation:

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After that program find the even values and then delete these values. After the operation new array will be displayed on output.

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Answer:

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