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bogdanovich [222]
3 years ago
12

Why has Unicode become the standard way of converting binary to text?

Computers and Technology
1 answer:
Zigmanuir [339]3 years ago
5 0
The standard for internal text coding system is called Unicode (Encoding, representation and handling text). Nearly every operating system uses this. The basic reason for choosing Unicode for converting binary to text is because this is compatible with almost every language in the world making it popular. Although there are other ways to do this, Unicode stays popular of all.
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Code works, need help writing header file.
vovikov84 [41]

Answer:

kbfbnmcj hbx nmjn mz kjn  mj     hk hhhhhhgdhggbddddddbvgfcgfctrctgvhyubujbnjinkikmokimloplkmkojihgftrdesw3l,kmjinhuygtfrde4sw34drftyghuijokkkkkkkkkkkkkkkijjjjjjjkopl[[[[[koiju98hytttttttttttttuhiouhygdtrsgyuihkl;;ihugydtrcfbhjklioutrcgfbhjkloiuytdrhjukliuytrdgfhjkiouy7gfhjklpoiudtrkhjl;[poiughjbnm,;lkihuygtfvbnmjkiuytfgchvbnmkjlhuygfdc vbnhnjkygtdsfxghbjkiuhytrdgfhjuioytrdfgchbnmkloiuytfdfghjiouyyyyyyhjkpoi8u97t6rdghjuo0987hjklpo0i98uyghjbkopiuyhjbnkploiuhyjbnmkloiuhygbvjnmk,l;oipuhgbjkn mjhuygfvbnhjkiouuuukjopipkjmloiujoklm;okjm,kloiuhjnmkiuhjbkiuygfhuji8yghyugfvhjkjiuhygfhuijokpiuyt67u89poiuhygftduhjklphuytfrdesdtgyuhi;lkyuuuugij'[;;;;;;;;;;;;;;;;;;;;okpuuuutyyyyy[poiudtryuiop[;oiudtrruiooooooooytg[pl;oiuhygtfffkjjjjhugvcfxkml,;kjnhbgvchbjnk,lijuhygfcbvh nmmpoiugyhjb,m ./[poiuhbjnm,k;lijuytfrcgbvhkjlipouuuujipouuuukliouygkjliubhjkml

Explanation:

7 0
3 years ago
Hello everyone. I need help. C programming. Create at least two more functions except the main () function to collect them.
Andrew [12]

Answer:

Explanation:

#include <iostream>

using namespace std;

int costdays(int);

int costhrs(int,int);

int main()

{

   int dd,hh,mm,tmph,tmpd,tmpm=0;

   int pcost,mcost=0;

   cout<<"Enter Parking time" << endl;

   cout<<"Hours: ";

   cin>>hh;

   cout<<"Minutes: ";

   cin>>mm;

   

   if (mm>60)

   {

       tmph=mm/60;

       hh+=tmph;

       mm-=(tmph*60);

   }

   if (hh>24)

   {

       tmpd=hh/24;

       dd+=tmpd;

       hh-=(tmpd*24);

   }

   if ((hh>4)&&(mm>0))

   {

       pcost+=costdays(1);

   }

   else

   {

       mcost=costhrs(hh,mm);

   }

   cout<<"Total time: ";

   if (dd>0)

   {

       cout<<dd<<"days ";

       pcost+=costdays(dd);

   }

   pcost+=mcost;

   cout<<hh<<"h "<<mm<<"mins"<<endl;

   cout<<"Total Cost :"<<pcost<<"Won";

   return 0;

}

int costdays(int dd)

{

   return(dd*25000);

}

int costhrs(int hh,int mm)

{

   int tmpm, tmp=0;

   tmp=(hh*6)*1000;

   tmp+=(mm/10)*1000;

   tmpm=mm-((mm/10)*10);

   if (tmpm>0)

   {

       tmp+=1000;

   }

   return(tmp);

}

3 0
3 years ago
Consider the following code segment.
mafiozo [28]

Answer:

for (int h = k; h >= 0; h--)

Explanation:

From the list of given options, option C answers the question.

In the outer loop

Initially, k = 0

In the inner loop,

h = k = 0

The value of h will be printed once because h>=0  means 0>=0 and this implies once

To the outer loop

k = 1

The inner loop will always assume value of k;

So,

h = 1

This will be printed twice because of the condition h>=0  means 1>=0.

Since 1 and 0 are >=0; 1 will be printed twice

To the outer loop

k = 2

The inner loop

h = 2

This will be printed thrice because of the condition h>=0  means 2>=0.

Since 2, 1 and 0 are >=0; 2 will be printed thrice

To the outer loop

k = 3

The inner loop

h = 3

This will be printed four times because of the condition h>=0  means 3>=0.

Since 3, 2, 1 and 0 are >=0; 3 will be printed four times

7 0
3 years ago
Print "Censored" if userInput contains the word "darn", else print userInput. End with newline. Ex: If userInput is "That darn c
barxatty [35]

Answer:

#include <string>

#include <iostream>

using namespace std;

int main() {

string userInput;

getline(cin, userInput);

// Here, an integer variable is declared to find that the user entered string consist of word darn or not

int isPresent = userInput.find("darn");

if (isPresent > 0){

cout << "Censored" << endl;

// Solution starts here

else

{

cout << userInput << endl;

}

// End of solution

return 0;

}

// End of Program

The proposed solution added an else statement to the code

This will enable the program to print the userInput if userInput doesn't contain the word darn

6 0
3 years ago
Question 2 (2 points)
swat32

I beleive all of the above for question one

4 0
3 years ago
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