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Sholpan [36]
2 years ago
5

Writers in our networked world enhance their credibility by

Computers and Technology
1 answer:
yaroslaw [1]2 years ago
5 0

Answer:

b

Explanation:

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Just a quick question, how do you set something == to char and int in an if statement (java)
Troyanec [42]

This is the requested code in java.

public class <em>CharTest</em> {

   public static String checkCharacter(String text, int index) {

       if (0 <= index && index <= text.length()) {

           char ch = text.charAt(index);

           if (Character.isLetter(ch)) {

               return ch + " is a letter";

           } else if (Character.isDigit(ch)) {

               return ch + " is a digit";

           } else if (Character.isWhitespace(ch)) {

               return ch + " is a whitespace character";

           } else {

               return ch + " is an unknown type of character";

           }

       } else {

           return "index " + index.toString() + " is out of range";

       } // <em>end if</em>

   } // <em>end function checkChar()</em>

   public static void <em>main</em>(String[] args) {

       // <em>Test the three samples from the specification.</em>

       System.out.println(<em>checkCharacter</em>("happy birthday", 2));

       System.out.println(<em>checkCharacter</em>("happy birthday", 5));

       System.out.println(<em>checkCharacter</em>("happy birthday 2 you", 15));

   } // <em>end function main()</em>

} // <em>end class CharTest</em>

The function <em>checkcharacter</em>(text, index) returns a string value describing the kind of character found at the position in text specified by index; whether it was a letter, digit, whitespace, or an unknown kind of character.

How it does that is to make use of respective functions defined within the Character class in java. That is

  • isLetter(char) returns a bool specifying if the char parameter is a letter.
  • isDigit(char) returns a bool specifying if the char parameter is a digit.
  • isWhitespace(char) returns a bool specifying if the char parameter is a whitespace character.

It calls these functions in an if statement. These else part of the if statement is then executed if the character is neither a <em>letter</em>, <em>digit</em>, or <em>whitespace</em>.

Finally, the function main() calls <em>checkCharacter</em>() three times to test the function and return the results to the console.

Another example of a java program on characters is found in the link below

brainly.com/question/15061607

3 0
2 years ago
#Draw patterns using fibonacci
agasfer [191]

Answer:

yidRYoYORoydzyrozoyrsoyraoyeahplhDogdaoyeayeoayorahdlHDLgksOyeayeazhldhdlHkdOyaoyaerhaahdLhKDoywaeya9yaeya9eeY9ey9

7 0
3 years ago
5)What are the differences in the function calls between the four member functions of the Shape class below?void Shape::member(S
Stells [14]

Answer:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

Explanation:

void Shape :: member ( Shape s1, Shape s2 ) ; // pass by value

The s1 and s2 objects are passed by value as there is no * or & sign with them. If any change is made to s1 or s2 object, there will not be any change to the original object.

void Shape :: member ( Shape *s1, Shape *s2 ) ; // pass by pointer

The s1 and s2 objects are passed by pointer as there is a * sign and not & sign with them. If any change is made to s1 or s2 object, there will be a change to the original object.

void Shape :: member ( Shape& s1, Shape& s2 ) const ; // pass by reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. If any change is made to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) ; // pass by const reference

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. The major change is the usage of const keyword here. Const keyword restricts us so we cannot make any change to s1 or s2 object.

void Shape :: member ( const Shape& s1, const Shape& s2 ) const ; // plus the function is const

The s1 and s2 objects are passed by reference  as there is a & sign and not * sign with them. const keyword restricts us so we cannot make any change to s1 or s2 object as well as the Shape function itself.

5 0
3 years ago
What is a dbms in full
Alika [10]

database management system

5 0
3 years ago
Using SQL
lesantik [10]

Answer:

Following are the code to this question:

/*using the select statement, that selects column name from the table blog.posts */  

SELECT blog.posts.user_id, blog.posts.body, users.name/*column name user_id, body, name*/  

FROM blog.posts/* use table name blog.posts*/

RIGHT OUTER JOIN users ON blog.posts.user_id = users.id;/*use right join that connect table through user_id*/

Explanation:

In the structured query language, RIGHT JOIN  is used to recovers from both the right side of the table both numbers, although the left table has no sets. It also ensures that even if the 0 (null) documents are linked inside this left table, its entry will always return the outcome row, but still, the number of columns from its left table will be NULL.

In the above-given right join code, the select statements used that selects the column names "user_id, body, and the name" from the table "blog. posts" and use the right join syntax to connect the table through the id.    

7 0
3 years ago
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