CODEWITHSUNDEEP
java
vinay
vinay

Reputation: 171

Can anyone explain this code for me?

public int sumOfDigits(int num){
   return num == 0 ? 0 : num % 10 + sumOfDigits(num/10);
}

Upvotes: 0

Views: 122

Answers (7)

Cebu CM Solutions
Cebu CM Solutions

Reputation: 111

That's a Java ternary operator, but in most cases it is not used since if-else statements are much more readable, which can also do the same function ternary operators can do. But if you would like to make shorter code lines, then this is preferred.

Upvotes: 0

Dinesh Peiris
Dinesh Peiris

Reputation: 56

This method calculates sum of the numbers representing base 10 integer numbers. The method is called recursively with % operator to separate digits in the number and then add them to form the result.

Eg - 234 into 2 + 3 + 4 = 9

However, if you pass a integer in base other than base 10, the method still gives the result for base 10

Eg sumOfDigits(0b1011010) -> answer is 9 for 90 in base 10 sumOfDigits(0x532) -> answer is 7 for 1330 in base 10

Following is the change for binary and octal values to get the sum of the representing numbers

return num == 0 ? 0 : num % 2 + sumOfDigits(num / 2);

return num == 0 ? 0 : num % 8 + sumOfDigits(num / 8);

Upvotes: 0

jdphenix
jdphenix

Reputation: 15445

That is the Java ternary operator. It is similar to ternary operators in other languages. In this case: 

return <boolean expression> ? <if true> : <if false>;

You have posted a recursive algorithm to calculate the sum of digits in an integer.

Two equivalent methods are posted below: 

public int sumOfDigitsIf(int num) { 
    if (num == 0) { 
        return 0; 
    } else {
        return num % 10 + sumOfDigitsIf(num / 10);
    }
}

public int sumOfDigitsLoop(int num) {  // useful to examine if you aren't used to recursive algorithms
    int value = 0; 
    while (num > 0) { 
        value += num % 10;
        num /= 10;
    }
    return value;
}

Upvotes: 1

Darshan Lila
Darshan Lila

Reputation: 5868

return num == 0 ? 0 : num % 10 + sumOfDigits(num/10);

Is equivalent to:

if(num==0)
{
 return 0;
}
else
{
 return num % 10 + sumOfDigits(num/10); // this is recursive call. It gets called until num is reduce to 0
}

This basically calulates as following:
If num=768 result= 7+6+8.

Upvotes: 1

Straw1239
Straw1239

Reputation: 689

This function calculates the sum of the digits of the number recursively. First, if the number is 0, it returns 0, as the sum of the digits of 0 is 0. Then, it calculates the digit in the 1s place with num % 10, and adds that to the sum of the digits of the remaining places, num / 10 is the remaining places, and it calls sumOfDigits() on that number.

Upvotes: 0

SpringLearner
SpringLearner

Reputation: 13854

sumOfDigits is a method which accepts int argument and the return type is also int. If 0 is passed then it will return 0 other wise it will return num % 10 + sumOfDigits(num/10)

If a number other than 0 is passed this will continue as long as entered number becomes 0

Upvotes: 1

Mureinik
Mureinik

Reputation: 312259

This is a recursive calculation of the sum of digits - as long as you haven't consumed all the digits, it adds the last one with the sum of all the previous ones.

Upvotes: 3

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