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kramer
2 years ago
9

"What technology is used to assure that the logical block addressing on a solid state drive does not always address the same phy

sical blocks, in order to distribute write operations?"
Computers and Technology
1 answer:
elena-14-01-66 [18.8K]2 years ago
3 0

Answer:

Wear leveling is the correct answer of this question.

Explanation:

Wear leveling is the technique used to ensure the same physical blocks are not necessarily addressed by the logical block addressing on a solid state drive. In this way, distribution of the writing operations is the right order.

Wear leveling is a method used in consistent-state storage (Sd cards) and USB hard drives, and in process-change memory, to enhance the shelf life of other kinds of acrylic disk storage devices.The idea is to allocate coding on all sheets of a Hdd literally equally, so that they wear equally.

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Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
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Answer:

<h2>Function 1:</h2>

#include <stdio.h> //for using input output functions

// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter

void PrintPopcornTime(int bagOunces){

if (bagOunces < 3){ //if value of bagOunces is less than 3

 printf("Too small"); //displays Too small message in output

 printf("\n"); } //prints a new line

//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10

else if (bagOunces > 10){

    printf("Too large"); //displays the message:  Too large in output

    printf("\n"); //prints a new line }

/*the following else  part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */

else {

/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */

    //int result;

    //result = bagOunces * 6;

    //printf("%d",result);

 printf("%d",bagOunces * 6);  /multiplies value of bagOunces  to 6

 printf(" seconds");

// seconds is followed with the value of bagOunces * 6

 printf("\n"); }} //prints a new line

int main(){ //start of main() function body

int userOunces; //declares integer variable userOunces

scanf("%d", &userOunces); //reads input value of userOunces

PrintPopcornTime(userOunces);

//calls PrintPopcornTime function passing the value in userOunces

return 0; }

Explanation:

<h2>Function 2:  </h2>

#include <stdio.h> //header file to use input output functions

// start of the function PrintShampooInstructions body having integer variable numCycles as parameter

void PrintShampooInstructions(int numCycles){

if(numCycles < 1){

//if conditions checks value of numCycles is less than 1 or not

printf("Too few."); //prints Too few in output if the above condition is true

printf("\n"); } //prints a new line

//else if part is executed when the if condition is false and else if  checks //value of numCycles is greater than 4 or not

else if(numCycles > 4){

//prints Too many in output if the above condition is true

printf("Too many.");

printf("\n"); } //prints a new line

//else part is executed when the if and else if conditions are false

else{

//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done

for(int N = 1; N <= numCycles; N++){

printf("%d",N);

printf(": Lather and rinse. \n");}

printf("Done.");

printf("\n");} }

int main() //start of the main() function body

{    int userCycles; //declares integer variable userCycles

   scanf("%d", &userCycles); //reads the input value into userCycles

   PrintShampooInstructions(userCycles);

//calls PrintShampooInstructions function passing the value in userCycles

   return 0;}

I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN  which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by

printf(": Lather and rinse. \n"); which is followed by printf("Done.");

So at first iteration:

printf("%d",N); prints 1 as the value of N is 1

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

1: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.

Now at second iteration:

The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles  which means 2=2. So the program control enters the body of loop.

printf("Done."); prints Done after the above two lines.

printf("%d",N); prints 2 as the value of N is 2

printf(": Lather and rinse. \n");  prints : Lather and rinse and prints a new line \n.

As a whole this line is printed on the screen:

2: Lather and rinse.

Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.

The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles  which means 3>2. So the loop breaks.

Now the next statement is:

printf("Done."); which prints Done on the screen.

So as a whole the following output is displayed on the screen:

1: Lather and rinse.

2: Lather and rinse.

Done.

The programs along with their outputs are attached.

6 0
2 years ago
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