Ask question

Ask question

All categories

3 years ago

10

Write a family database program. Create a class to represent a person and to store references to the person’s mother, father, an

d any children the person has. Read a file of names to initialize the name and parent–child relationships of each Person. (You might wish to create a file representing your own family tree.) Store the overall list of Persons as an ArrayList. Write an overall main user interface that asks for a name and prints the maternal and paternal family line for that person.

Computers and Technology

1 answer:

Gennadij [26K]3 years ago

8 0

Do you go to BASIS?

Sorry I don't have an answer for you, but we have the same assignment in our AP Comp sci class.

Just wondering.

You might be interested in

output device is any peripheral to provide data and control signal to ab information processing system

nirvana33 [79]

Answer:

Its false ita not "output" its "input"

7 0

3 years ago

The view of a presentation can be changed on the View tab or by selecting the appropriate icon on the _____.

ELEN [110]

The answer is Status bar

Hope my answer Helps! :)

7 0

3 years ago

Read 2 more answers

Write a program, named sortlist, that reads three to four integers from the command line arguments and returns the sorted list o

yawa3891 [41]

Answer:

In C++:

void sortlist(char nums[],int charlent){

int Myarr[charlent];

const char s[4] = " ";

char* tok;

tok = strtok(nums, s);

int i = 0;

while (tok != 0) {

int y = atoi(tok);

Myarr[i] = y;

tok = strtok(0, s);

i++;

}

int a;

for (int i = 0; i < charlent; ++i) {

for (int j = i + 1; j < charlent; ++j) {

if (Myarr[i] > Myarr[j]) {

a = Myarr[i];

Myarr[i] = Myarr[j];

Myarr[j] = a;

} } }

for(int j = 0;j<charlent;j++){ printf(" %d",Myarr[j]); }

}

Explanation:

This line defines the sortlist function. It receives chararray and its length as arguments

void sortlist(char nums[],int charlent){

This declares an array

int Myarr[len];

This declares a constant char s and also initializes it to space

const char s[4] = " ";

This declares a token as a char pointer

char* tok;

This line gets the first token from the char

tok = strtok(nums, s);

This initializes variable i to 0

int i = 0;

The following while loop passes converts each token to integer and then passes the tokens to the array

<em> while (tok != 0) { </em>

<em> int y = atoi(tok); </em><em>-> Convert token to integer</em><em> </em>

<em> Myarr[i] = y; </em><em>-> Pass token to array</em><em> </em>

<em> tok = strtok(0, s); </em><em>-> Read token</em><em> </em>

<em> i++; </em>

<em> } </em>

Next, is to sort the list.

int a;

This iterates through the list

for (int i = 0; i < charlent; ++i) {

This iterates through every other elements of the list

for (int j = i + 1; j < charlent; ++j) {

This condition checks if the current element is greater than next element

if (Myarr[i] > Myarr[j]) {

If true, swap both elements

<em> a = Myarr[i]; </em>

<em> Myarr[i] = Myarr[j]; </em>

<em> Myarr[j] = a; </em>

} } }

The following iteration prints the sorted array

<em>for(int j = 0;j<charlent;j++){ printf(" %d",Myarr[j]); } </em>

}

<em>See attachment for illustration of how to call the function from main</em>

7 0

3 years ago

(tco 5) how many 4-bit parallel adders are required to add the numbers 5,345 and 3,892? explain how you got your answer.

Igoryamba

The sum is 9237. To express this as binary requires log(9237)/log(2) bits ≈ 13.2 bits, rounded up at least 14 bits. (You can check: 2^13 is not enough, 2^14 is enough)

So you need four 4-bit adders, giving you 16 bits resolution.

7 0

3 years ago

(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-

I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit memory to add in 2 location:

$\to \frac{0}{1} =2^1 \ (\frac{m}{m} \ location)$

The address of 2-bit memory to add in 4 location:

$\to \frac{\frac{00}{01}}{\frac{10}{11}} =2^2 \ (\frac{m}{m} \ location)$

similarly,

Complete 'n'-bit memory address' location number is = $2^n.$ Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

$= 2^n \\\\ = 2^{12} \\\\ = 4096$

In point b:

$\to Let \ Mega= 10^6$

$=10^3\times 10^3\\\\= 2^{10} \times 2^{10}$

So,

$\to 2 \ Mega =2 \times 2^{20}$

$= 2^1 \times 2^{20}\\\\= 2^{21}$

The memory position for ' $2^n$ ' could be 'n' m bits'

It can use $2^{21}$ bits to address the memory location of 21.

That is to say, the 2-mega-location memory needs '21' bits.

Memory Length = 21 bit Address

In point c:

$i^{th}$ element array addresses are given by:

$\to address [i] = B+w \times (i-LB)$

$_{where}, \\\\B = \text {Base address}\\w= \text{size of the element}\\L B = \text{lower array bound}$

$\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address = 10$

$\to address [10] = \$ 52 + 4 \times (10-0)\\$

$= \$ 52 + 40 \ bytes\\$

1 term is 4 bytes in 'MIPS,' that is:

$= \$ 52 + 10 \ words\\\\ = \$ 512$

In point d:

$\to base \ address = \$ t 5$

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are < 4 byte

Calculating address:

$\to address [5] = \$ t5 + 4 \times (5-0)\\$

$= \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20 \ bytes \\\\= \$ t5 + 5 \ words \\\\= \$ t 10 \ words \\\\$

3 0

3 years ago