Truncating integer using string manipulation

Question

I have a class with a data member that needs to be rounded up to a 2 digit integer, irrespective of the number of the input digits.

For example:

roundUpto2digit(12356463) == 12 
roundUpto2digit(12547984) == 13 // The 5 rounds the 12 up to 13.

Currently my code looks like:

int roundUpto2digit(int cents){
    // convert cents to string
    string truncatedValue = to_string(cents);
    // take first two elements corresponding to the Most Sign. Bits
    // convert char to int, by -'0', multiply the first by 10 and sum the second 
    int totalsum =  int(truncatedValue[0]-'0')*10 + int(truncatedValue[1]-'0');
    // if the third element greater the five, increment the sum by one
    if (truncatedValue[2]>=5) totalsum++; 
    return totalsum;
} 

How can this be made less ugly?

Answer 1

If I understood your requirements correctly then it might be like this:

int roundUpto2digit(int cents) {
  if (cents < 100)
    return cents < 10 ? cents * 10 : cents;
  while ((cents + 5) >= 1000)
    cents /= 10;
  return (cents + 5) / 10;
}

The test:

#include <stdio.h>

void
test(int i) {
    printf("%d -> %d\n", i, roundUpto2digit(i));
}

int
main() {
    test(0);
    test(1);
    test(5);
    test(9);
    test(10);
    test(49);
    test(50);
    test(94);
    test(95);
    test(99);
    test(100);
    test(104);
    test(105);
    test(994);
    test(995);
    test(999);
    test(1000);
    test(1040);
    test(1050);
    return 0;
}

The result:

0 -> 0
1 -> 10
5 -> 50
9 -> 90
10 -> 10
49 -> 49
50 -> 50
94 -> 94
95 -> 95
99 -> 99
100 -> 10
104 -> 10
105 -> 11
994 -> 99
995 -> 10
999 -> 10
1000 -> 10
1040 -> 10
1050 -> 11

It is uncertain if [1, 9] range should map into [10, 90] (2 digits) or [1, 9] (1 digit). I could fix it, if the later case is true.

Answer 2

With some math you can avoid the string conversion.

First take the log10 of the input:

double lg10 = log10(input);

The integral part of that will contain the exponent while the fractional part will contain the digits of the mantissa.

Putting just the fractional part in the exponent of 10^exponent will create a number between 1 and 9.999 which is scaled by an exact power of ten. So we can multiply that number by 10 and round it for the result:

double frac = lg10 - floor(lg10); //note round to -inf to get the integral part
return (int)round(10*pow(10, frac));

This can overflow from 99.9 to 100 so we need to catch that.

Putting it all together results in:

int roundUpto2digit(int cents){
    if(cents == 0)
        return 0;

    double lg10 = log10(cents);
    double frac = lg10 - floor(lg10); 

    int result = (int)round(10*pow(10, frac));
    return result==100 ? 10 : result;
}

Answer 3

I don't know about making the code more elegant, but a faster way is to use a chain of ifs:

int mostSignificantThreeDigits(int value){
    if (value < 1000) return cents;
    if (value < 10000) return cents/10;
    if (value < 100000) return cents/100;
    // etc.
}

int nearestMultipleOf10Factor(int value){
    return (value+5)/10;
}

int truncatedToTwoDigits(int value){
    return nearestMultipleOf10Factor(mostSignificantThreeDigits(value));
}

It's also possible to use a loop for mostSignificantThreeDigits():

int mostSignificantThreeDigits(int value){
    while (value > 1000) {
      value /= 10;
    }
    return value;
}

but I suspect it's slower due to the repeated divisions.

int mostSignificantThreeDigits(int value){
    int factor = 1000;
    int divisor = 1;
    while (value > factor) {
      factor *= 10;
      divisor *= 10;
    }
    return value / divisor;
}

might be faster if integer multiplication is way faster than integer division.