Get the Sum of All Digits in a Numeric String

Sum all the digits of a number Javascript

Basically you have two methods to get the sum of all parts of an integer number.

• With numerical operations

  Take the number and build the remainder of ten and add that. Then take the integer part of the division of the number by 10. Proceed.

var value = 2568,
    sum = 0;

while (value) {
    sum += value % 10;
    value = Math.floor(value / 10);
}

console.log(sum);

Find Sum From String of Numbers

Here is an easy way to do it.

String string =
        "This12is  mix of 3 characters 45spaces and numbers 67";

• initialize sum to 0;
• loop thru all the characters
• Check that the character is a digit
• if so, subtract '0' and sum the result.

int sum = 0;
for (int i = 0; i < string.length(); i++) {
    char c = string.charAt(i);
    if (Character.isDigit(c)) {
        sum = sum + c - '0';
    }
}

Prints

28

About subtracting '0'

ASCII and Unicode characters are binary values that are encoded for display on some output device. This includes digits. However, digits are not in proper format for numeric computations so they need to be converted to their numeric binary format. The easiest way is to subtract decimal 48 (which is the encoded representation for the digit 0) from each digit prior to summing them. For larger numbers like "29929" Integer.parseInt() can be used. Internally it does similar processing of the individual digits as previously described.

Display the sum of digits in a number entered by the user

This is a very common mistake. char is really just a number - the encoding value of the character represented by the char. When you do Convert.ToInt32 on it, it sees the char as a number and says "alright let's just convert this number to 32 bits and return!" instead of trying to parse the character.

"Wait, where have I used a char in my code?" you might ask. Well, here:

Convert.ToInt32(num[count]) // 'num[count]' evaluates to 'char'

To fix this, you need to convert the char to a string:

nums[count] = Convert.ToInt32(num[count].ToString());
                              ^^^^^^^^^^^^^^^^^^^^^

Now you are calling a different overload of the ToInt32 method, which actually tries to parse the string.

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