Given Information : The given code is: // define structure square point_color { // variable declaration int c; int m; int y; int k; }; // declare structure array struct point_color square[16][16]; int i,j; // traverse through the array for(i=0;i<16;i++) { //traverse through elements for(j=0;j<16;j++) { //square all elements of the 2-D array square[i][j].c=0; square[i][j].m=0; square[i][j].y=1; square[i][j].k=0; } } Write hit: If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty. If the state of information is in valid state then it executes a write-through operation. It then updates the memory and block and changes its blocked state to reserved state. Write miss : A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Question

Want to see the full answer?

Answer 1

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

D. S. Malik, Data Structures Using C++, 2nd Edition, 2010

Methods (Ch6) - Review 1. (The MyRoot method) Below is a manual implementation of the Math.sqrt() method in Java. There are two methods, method #1 which calculates the square root for positive integers, and method #2, which calculates the square root of positive doubles (also works for integers). public class SquareRoot { public static void main(String[] args) { } // implement a loop of your choice here // Method that calculates the square root of integer variables public static double myRoot(int number) { double root; root=number/2; double root old; do { root old root; root (root_old+number/root_old)/2; } while (Math.abs(root_old-root)>1.8E-6); return root; } // Method that calculates the square root of double variables public static double myRoot(double number) { double root; root number/2; double root_old; do { root old root; root (root_old+number/root_old)/2; while (Math.abs (root_old-root)>1.0E-6); return root; } } Program-it-Yourself: In the main method, create a program that…

I would like to know the main features about the following 3 key concepts:1. Backup Domain Controller (BDC)2. Access Control List (ACL)3. Dynamic Memory

Answer 2

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 6

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 7

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 8

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 9

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 10

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 11

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

for(i=0;i<16;i++)

{

//traverse through elements

for(j=0;j<16;j++)

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 6.1 - Prob. 6.1PP Ch. 6.1 - Prob. 6.2PP Ch. 6.1 - Prob. 6.3PP Ch. 6.1 - Prob. 6.4PP Ch. 6.1 - Prob. 6.5PP Ch. 6.1 - Prob. 6.6PP Ch. 6.2 - Prob. 6.7PP Ch. 6.2 - Prob. 6.8PP Ch. 6.4 - Prob. 6.9PP Ch. 6.4 - Prob. 6.10PP

Want to see the full answer?

Chapter 6 Solutions