Question

What will the code shown below print to the console? public class Cascade public static void print(int arg){ for(int i=0; i<arg; i++){ System.out.print(""); } System.out.println(""); if(arg > 1){ print(arg-1); } } public static void main(String[] args) { print(4); } } E B IV AA- IES XX, SE V GT 12pt Paragraph -

Answer 1

class Cascade{

public static void print(int arg)

{

   for(int i=0;i<arg;i++)

   {

     System.out.print("-");

   }

   System.out.println(">");

   if(arg >1)

   {

     print(arg-1);

   }

}

public static void main(String[] args) {

print(4);

}

}

====================

//Output of the above code is

---->
--->
-->
->

Explaination

From main print method is called with argument 4.

Inside method print

there is for loop which loops from i=0 till i< 4 , that means it loops 4 times printing char "-" 4 times continously as like this ----, after for loop there is print statement which prints > just after 4 characters of "-" like this ---->, next statement is if statement which checks if arg is greater than 1, here its arg-1 ,which is 3 . So again print method is called , but this time with arg 3 , again same lines executed for loop and printing char ">", this time isntead of 4 times , for loops will get executed 3 times, hence we get output as "--->", again if statement executed , this time print called with arg-1 ie 3-1=2 , Output of calling print with arg 2 is --> , next time print will be called with print(1) , this time we get output as -> , and if statement arg >1 is false as arg has become 1 after calling subsequent method call. So recursive function will return to main and we get output as mentioned above.

Conclusion: print is called recursively till arg is greater than 1.