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Wittaler [7]
3 years ago
6

TV show information can either keep track of the number of viewers per show or the name of a show. Information is kept in a file

(see sample files below). If the first line of the file says Show, we should store the names of the shows. If the first line of the file says Viewer, we should store the number of viewers per show.
Create a class to complete the program below.
int main(int argc, char** argv) Tv_show_info Tv_show_info show_names(argv[2]);
int n=viewers.info.at(0);
std::string s=show_names.info.at(0);

Sample input file 2: Viewer Sample input file 1: Show Stranger Things Game of Thrones Black Mirror 34 100

its not provided, we are to recreate it. but this is what i made..

template
class Tv_show_info
{
public:
// template
Tv_show_info(string & file):
filename{file}
{
ifstream myfile {filename};
vectorinfo;
T lines;
while (getline(myfile,lines))
{
info.push_back(lines);
}
}
private:
string filename;
};

Computers and Technology
1 answer:
Elza [17]3 years ago
3 0

Answer:

See explaination for Program source code.

Explanation:

The program source code below.

#include <iostream>

#include <fstream>

#include <vector>

using namespace std;

class Show

{

private:

string title;

int viewers;

public:

Show()

{

this->title = "";

this->viewers = 0;

}

Show(string title, int viewers)

{

this->title = title;

this->viewers = viewers;

}

string getTitle(){ return this->title; }

int getViewers(){ return this->viewers; }

};

int main()

{

vector<Show> shows;

vector<string> titles;

vector<int> viewers;

ifstream inFile1("file1.txt");

ifstream inFile2("file2.txt");

string line1, line2;

if(!inFile1.is_open() || !inFile2.is_open())

{

cout << "Either of the files could not be found!\n";

exit(0);

}

getline(inFile1, line1);

if(line1.compare("Show") == 0)

{

while(getline(inFile1, line1))

{

titles.push_back(line1);

}

inFile1.close();

}

getline(inFile2, line2);

if(line2.compare("Viewer") == 0)

{

while(getline(inFile2, line2))

{

viewers.push_back(stod(line2));

}

inFile2.close();

}

for(int i = 0; i < titles.size(); i++)

{

shows.push_back(Show(titles[i], viewers[i]));

}

// display all the show details

cout << "\nALL SHOWS:\n----------\n";

for(Show show : shows)

{

cout << "Title: " << show.getTitle() << ", Number of viewers: " << show.getViewers() << endl;

}

cout << endl;

return 0;

}

See attachment for output

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Answer:

Following are the statement is given below

if(updateDirection ==1) // check condition

{

++numUsers; // increments the value

}

else

{

--numUsers; // decrement the value

}

Explanation:

Following are the description of statement

  • Check the condition in the if block .If the "updateDirection" variable is 1 then then control moves to the if block otherwise control moves to the else block statement.
  • In the if  block the statement is "++numUsers" it means it increment the value by 1 .
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4 years ago
Read 2 more answers
Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument
timofeeve [1]

Answer:

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

Explanation:

For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

7 0
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Answer:

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Verizon [17]

Answer:

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Explanation:

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  • But the other option is not correct because the option A states that the change the name of the variable which is not a result of anything.
  • And the option D states that the code will be retyped again which is also not justify anything.

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