Computer Science, asked by vinaykumar891234, 2 months ago

interface LangFunction
{
void callMe();
}
class HackerEarth{
public static void main(String args[])
{
String str =
"Hacking";
LangFunction if () -> System.out.println(str+"java");
str="ay";
Lf.callme();
}
}​

Answers

Answered by Anonymous
0

Answer:

j,b,c are the lengths of the sides as shown in

the figure. Write the following ratios,

(i) sin X (ii) tan Z (iii) cos X (iv) tan X.

Answer

1)sin X=opposite side ofangle X

Hypotenuse

Answered by dreamrob
3

Answer:

Given program:

interface LangFunction

{

 void callMe();

}

class HackerEarth

{

    public static void main(String []args)

    {

       String str = "Hacking";

       LangFunction Lf = ()-> System.out.println(str+ "java");

       str = "ay";

       Lf.callMe();

    }

}

If we will execute the above program, then we will get an error.

error: local variables referenced from a lambda expression must be final or effectively final

       LangFunction Lf = ()-> System.out.println(str+ "java");

This is because all the attributes in an interface are by default public, static and final.

So, we need to make str variable final.

After some changes in the program:

interface LangFunction

{

 void callMe();

}

class HackerEarth

{

    public static void main(String []args)

    {

       final String str = "Hacking";

       LangFunction Lf = ()-> System.out.println(str+ "java");

       str = "ay";

       Lf.callMe();

    }

}

Although we have made the str variable final. Still we are getting error while compilation.

error: cannot assign a value to final variable str

       str = "ay";

This is because we are trying to change the value of a final variable. We cannot do so.

To remove the error, just comment out that statement.

After these changes:

interface LangFunction

{

 void callMe();

}

class HackerEarth

{

    public static void main(String []args)

    {

       final String str = "Hacking";

       LangFunction Lf = ()-> System.out.println(str+ "java");

       //str = "ay";

       Lf.callMe();

    }

}

The above program will execute properly without any error and we will get an output.

Output: Hackingjava

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