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Masja [62]
3 years ago
12

PLS HELP MEE!!!

Computers and Technology
1 answer:
kramer3 years ago
4 0

Answer:

starting with lowercase letters, using uppercase letters for the first letter in a new word

Explanation:

starting with a dollar sign, using lowercase for the remaining parts of each word

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What is the difference between compliled and intebrated language?
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Answer:  

A compiled language is a programming language whose implementations are typically compilers and not interpreters. In this language, once the program is compiled it is expressed in the instructions of the target machine. There are at least two steps to get from source code to execution. While, an interpreted language is a programming language whose implementations execute instructions directly and freely, without previously compiling a program into machine-language instructions. While in this language, the instructions are not directly executed by the target machine. There is only one step to get from source code to execution.

I hope this helps.

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From the television industry's point of view, which of the following would be the most desirable viewer?
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A 50-year-old African American man
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Using the C language, write a function that accepts two parameters: a string of characters and a single character. The function
Sav [38]

Answer:

#include <stdio.h>

void interchangeCase(char phrase[],char c){

  for(int i=0;phrase[i]!='\0';i++){

      if(phrase[i]==c){

          if(phrase[i]>='A' && phrase[i]<='Z')

              phrase[i]+=32;

          else

              phrase[i]-=32;      

      }

  }

}

int main(){

  char c1[]="Eevee";

  interchangeCase(c1,'e');

  printf("%s\n",c1);

  char c2[]="Eevee";

  interchangeCase(c2,'E');

  printf("%s\n",c2);    

}

Explanation:

  • Create a function called interchangeCase that takes the phrase and c as parameters.
  • Run a for loop that runs until the end of phrase and check whether the selected character is found or not using an if statement.
  • If the character is upper-case alphabet, change it to lower-case alphabet and otherwise do the vice versa.
  • Inside the main function, test the program and display the results.
8 0
3 years ago
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

7 0
3 years ago
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