I am new to Java , There is an error in my recursive function , when I pass value 10 to the function it behaves normally and print 10 to 0 .but when passing 11 it breaks and printing value from -37663(instead of starting from 10 ) and ends at -47865 and stackOverFlow Error. pasting the code below .

public class Apna2 {

public static void main(String [] args ) {

    printNumber(11);
    


}   

public static void printNumber(int num ) {
    System.out.println(num);
    
    if(num==0)return;

    printNumber(num-2);
    
}

}

CodePudding user response：

When you pass 11, the sequence is 11, 9, 7, 5, 3, 1, -1, -3, .... Your base case if(num==0)return; is never executed. This is an infinite recursion. Your function call stack gets too big to fit into allowed memory space, hence you get a stack overflow error. This will happen for all odd numbers in your code.

CodePudding user response：

The recursive function would like to stop when the value reaches 0. Yet, if you have an odd number (like 11), then subtracting 2 from it repeatedly will result in the fullowing:

• 11 == 0 is false, therefore we call the method with 9
• 9 == 0 is false, therefore we call the method with 7
• 7 == 0 is false, therefore we call the method with 5
• 5 == 0 is false, therefore we call the method with 3
• 3 == 0 is false, therefore we call the method with 1
• 1 == 0 is false, therefore we call the method with -1
• ... and so on

You need to check whether your number is still positive and return if not:

public static void printNumber(int num ) {
    System.out.println(num);
    
    if(num<=0)return;

    printNumber(num-2);
    
}

CodePudding user response：

    public static void main(String [] args ) {

        printNumber(11);
        
    
    
    }   
    
    public static void printNumber(int num ) {

        if(num<0){num=0;}; //If number is below 0 set 0. ...
        System.out.println(num);
        if(num==0)return; //That this condition would be true.
        printNumber(num-2);
        
    }