Write a recursive function that returns the sum of the decimal digits of n, with the sign dropped so a negative n sums the digits of its magnitude.

• n is any int, anywhere from INT_MIN to INT_MAX. Return the sum of the decimal digits it is written with.
• Solve this with recursion, and let the number hand you the split: n % 10 is the last digit, n / 10 is everything written before it. So digit_sum(n) is that last digit plus digit_sum(n / 10).
• The base case is a number written with one digit, meaning n between -9 and 9. Return the digit and stop recursing.
• The sign is not part of the sum. digit_sum(-42) returns 6, exactly like digit_sum(42). Your return value is never negative.
• digit_sum(0) returns 0.
• n % 10 on a negative n gives a negative remainder: -42 % 10 is -2, since integer division in C truncates toward zero. Drop the sign of that one digit, which is always safe because a digit is at most 9, or accumulate the negative remainders and flip the total at the end.
• Do not turn a negative n positive first. -INT_MIN is 2147483648, one past the top of the int range, and signed overflow is undefined behavior: the standard says nothing about the result and the compiler is free to keep the value negative, wrap it, or delete the expression. A solution written around n = -n cannot reach the answer for INT_MIN.
• digit_sum(INT_MIN) returns 47 and digit_sum(INT_MAX) returns 46.
• The tests print your return value with %d followed by a newline.

Write a recursive function that returns the sum of the decimal digits of n, with the sign dropped so a negative n sums the digits of its magnitude.

• n is any int, anywhere from INT_MIN to INT_MAX. Return the sum of the decimal digits it is written with.
• Solve this with recursion, and let the number hand you the split: n % 10 is the last digit, n / 10 is everything written before it. So digit_sum(n) is that last digit plus digit_sum(n / 10).
• The base case is a number written with one digit, meaning n between -9 and 9. Return the digit and stop recursing.
• The sign is not part of the sum. digit_sum(-42) returns 6, exactly like digit_sum(42). Your return value is never negative.
• digit_sum(0) returns 0.
• n % 10 on a negative n gives a negative remainder: -42 % 10 is -2, since integer division in C truncates toward zero. Drop the sign of that one digit, which is always safe because a digit is at most 9, or accumulate the negative remainders and flip the total at the end.
• Do not turn a negative n positive first. -INT_MIN is 2147483648, one past the top of the int range, and signed overflow is undefined behavior: the standard says nothing about the result and the compiler is free to keep the value negative, wrap it, or delete the expression. A solution written around n = -n cannot reach the answer for INT_MIN.
• digit_sum(INT_MIN) returns 47 and digit_sum(INT_MAX) returns 46.
• The tests print your return value with %d followed by a newline.