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katovenus [111]
4 years ago
8

How many registers are there in the memory unit?

Computers and Technology
1 answer:
Sati [7]4 years ago
5 0

Answer:

Most modern CPU's have between 16 and 64 General Purpose Registers. Special Purpose Registers have special uses and are either nonprogrammable and internal to the CPU or accessed with special instructions by the programmer.

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11.
kodGreya [7K]

Answer:

The basic difference between finite and infinite is the number of times it runs and ends. The loop is basically a set of instructions that keeps repeating itself.

The finite loop ends after running for a finite times. This body of finite loop will stop executing after certain condition is reached. Hence the finite loop body keeps executing itself finite number of times.

An infinite loop keeps running and repeating itself endlessly.This loop never ends. This loop can be the result of an error in a program. For example when no stopping or exit condition is specified in the program.

Explanation:

Example of finite loop:

Lets take for loop as an example:

for(int i =0; i<=5; i++)

{ cout<<i<<endl; }

Now the loop starts from i=0

Next it enters the body of loop to execute the statement: cout<<i; which means that the value of i is displayed on the output screen.

This loop keeps executing until the value of i exceeds 5.

At first iteration 0 is printed on the output screen, at second iteration 1, at third iteration 2, at fourth iteration 3, fifth iteration 4, sixth iteration 5. After each of these iterations, the value of i is incremented by 1.

When 5 is printed, then at the next iteration the specified condition i.e. i<=5 gets false as the value of i now becomes 6 after incremented by 1.

So the loop stops running. So this means that loop ran for finite times and stopped after the a certain condition is reached. The output is:

0

1

2

3

4

5

Example of infinite loop:

Lets take while loop:

int i = 6;

    while (i >=5)

    {         cout<< i;

             i++;     }

In the above example, the loop will run infinite times. Here the value of i is initialized to 6. Then while loop is used which checks the condition which is the value of i is greater than or equal to 5. If the condition is true, the body of the loop starts executing which prints the value of i. Lets see what happens at each iteration:

1st iteration: i>=5 is True because i=6 and 6 is greater than 5. The program control enters the body of loop. The statement cout<<i prints 6. Next the value of i is incremented by 1 and it becomes 7.

2nd iteration: i>=5 is True because i=7 and 7 is greater than 5. The program control enters the body of loop. The statement cout<<i prints 7. Next the value of i is incremented by 1 and it becomes 8.

This loop will repeat itself infinite times and never stops as i will always have value greater than 5. So this is the example of infinite loop.

6 0
3 years ago
True or False? Wireless connections that use a higher frequency are faster but have a shorter range.
Rudiy27

Answer:

it's a True statement

6 0
3 years ago
Consider two different implementations of the same instruction set architecture. The instructions can be divided into four class
patriot [66]

Answer:

Processor P_2 is faster than P_1.

Part A:

For Processor P_1:

CPI=2.6

For Processor P_2:

CPI=2

Part B:

For Processor P_1:

CPU CLOCK CYClE=2.6*10^{6}

For Processor P_2:

CPU CLOCK CYClE=2*10^{6}

Explanation:

We will find the instructions executed of each class:

For Class A=1*10^{6} *10%=1*10^{5}

For Class B=1*10^{6} *20%=2*10^{5}

For Class C=1*10^{6} *50%=5*10^{5}

For Class D=1*10^{6} *20%=2*10^{5}

CPU CLOCK CYClE=\sum_{n=1}^{i}*CPI_i*C_i

Where:

C_i is the instructions of each class (Calculated above)

For Processor P_1:

CPU CLOCK CYClE=(1*1*10^{5})+ (2*2*10^{5})+ (3*5*10^{5})+ (3*2*10^{5})

CPU CLOCK CYClE=2.6*10^{6}

For Processor P_2:

CPU CLOCK CYClE=(2*1*10^{5})+ (2*2*10^{5})+ (2*5*10^{5})+ (2*2*10^{5})

CPU CLOCK CYClE=2*10^{6}

Now:

CPU TIME=CPU CLOCK CYClE/CYCLE RATE

For Processor P_1:

CPU TIME=\frac{2.6*10^{6} }{2.5*GHz} =1.04ms

For Processor P_2:

CPU TIME=\frac{2*10^{6} }{3*GHz} =666.67ms

Processor P_2 is faster than P_1.

Part A:

CPI=CPU CLOCK CYClE/Number of instructions

For Processor P_1:

CPI=\frac{2.6*10^{6}}{1*10^{6}}=2.6

For Processor P_2:

CPI=\frac{2*10^{6}}{1*10^{6}}=2

Part B:

For Processor P_1:

CPU CLOCK CYClE=(1*1*10^{5})+ (2*2*10^{5})+ (3*5*10^{5})+ (3*2*10^{5})

CPU CLOCK CYClE=2.6*10^{6}

For Processor P_2:

CPU CLOCK CYClE=(2*1*10^{5})+ (2*2*10^{5})+ (2*5*10^{5})+ (2*2*10^{5})

CPU CLOCK CYClE=2*10^{6}

4 0
3 years ago
Debbie would like to create a file that she can use as a starting point to create other files for her English class. What should
Citrus2011 [14]

Answer:

c. a template.

Explanation:

She should first create a Template so that she can use it as a starting point to create other files.

A template is file created with a blueprint with  a format that can be used for a group of documents.For example:- there are different templates present of the resume.

Hence the answer to this question is template.

4 0
4 years ago
What is the difference between a Is your Milling machine and grinding machine
Law Incorporation [45]

Answer:

one is used for milling and one is used for grinding

Explanation:

8 0
3 years ago
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