# Truncating integer using string manipulation

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++;
}


How can this be made less ugly?

• You've got a bug in your code. You forgot the singlequotes on if (truncatedValue[2]>='5') totalsum++;  – Snowbody Mar 31 '15 at 13:23
• What if you have less than 2 digits initially? E.g. roundUpto2digit(1)? – Aleksey Demakov Mar 31 '15 at 13:32
• @ Aleksey Demakov that is a valid question. A line checking for length and if statement including only the MSB will do it. Thanks! – Ziezi Mar 31 '15 at 13:38
• What should be the result of roundUpto2digit(999)? Following your code the result will be 100. But this is a 3-digit integer. Is this okay or should it be further truncated to 10 to make a 2-digit integer? – Aleksey Demakov Mar 31 '15 at 14:35
• @Aleksey Demakov one more error. Yes, it should always be 2-digit integer. – Ziezi Mar 31 '15 at 14:53

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.

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;
}


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.

• The original code is undefined for values in the [0, 99] range. There will be out of bounds sting access in this case. Also there is ambiguity for values [995, 999]. The result will be 100, a 3-digit value, which contradicts the task description to get just 2 digits. With your version the result is the same in this case. For the former case your version will return a single-digit value. For instance, the return value for argument 50 will be 5. This is again contradicts the requirement to return always 2 digits. – Aleksey Demakov Mar 31 '15 at 15:09