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valkas [14]
3 years ago
6

Copy the following code and run it. You should break it into the following 3 functions

Computers and Technology
1 answer:
Darya [45]3 years ago
6 0

Answer:

The solution code is as follows:

  1. #include <iostream>
  2. using namespace() std;
  3. const double PI = 3.14159;
  4. int getValidInput() {
  5.    int radius;
  6.    cout << "Type a -1 for radius to exit" << endl;
  7.    cout << "Enter the radius: ";
  8.    cin >> radius;
  9.    
  10.    while(radius < 0){
  11.        cout << "Type a -1 for radius to exit" << endl;
  12.        cout << "Enter the radius: ";
  13.        cin >> radius;
  14.    }
  15.    return radius;
  16. }
  17. double circleCalculations(int r){
  18.    
  19.        double circ = 2 * PI * r;
  20.        double area = PI * pow(r, 2);
  21.        return area;
  22. }
  23. void printResults(int r, double a){
  24.    
  25.    cout.setf(ios::fixed);
  26.    cout.precision(1);
  27.    cout << "A circle with radius " << r << " has a circumference of "
  28.    << circ << " and an area of " << a << endl << endl;  
  29. }
  30. int main()
  31. {
  32.    int radius;
  33.    double area;
  34.    double circ;
  35.    radius = getValidInput();
  36.    area = circleCalculations(radius);
  37.    printResults(radius, area);
  38. }

Explanation:

In this question, we can identify three main functional blocks from the original codes. We cut the block of codes that ask for user input radius and paste it into the new function getValidInput() (Line 6 - 9). However, we need to add the code in Line 11-15 to ensure only the radius with valid value will be returned.

Next, we can proceed to cut another block of code that calculate the circumference and area and paste it into function circleCalculations (Line 19 -24).

Next, we cut the code that display the radius and area and paste it to function printResults(Line 26-32).

At last, we call all the three functions in the main program (Line 39-41).

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The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

                                       CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

                    = 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x  10⁹ x 0.75 / 3 Ghz

                  = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

     Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

                = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that  CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore,   numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x  10⁹ / 0.75  = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall  that:

MIPS = Clock rate / CPI X 10⁶

  So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

        MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

  We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

                  MFLOPS₁ = 1777.6

                  MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

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