How to round off the totalcost (double) to nearest inetegrs - c++

Question

I have tried to search similar questions all over the net but non was useful for me.
The nearest I got to was "If the number before the .5 is odd round up, if even round down 13.5 turns into 14 but 12.5 turns into 12".
Coming to the question:
I simply need to calculate the total amount after a meal with the formula;
total amount = mealamount+ mealamount*tip% + mealamount *tax%
I came up with this piece of code (rough only)
#include<iostream>
#include<math.h>
#include <iomanip>
using namespace std;
int main () {
double mealCost;
double total=0;
double tipp,taxx=0;
cin>>mealCost;
int tip;
cin>>tip;
tipp=(tip*mealCost)/100;
int tax;
cin>>tax;
taxx=(tax*mealCost)/100;
total=mealCost+tipp+taxx;
cout << fixed << showpoint;
cout << setprecision(2);
cout<<total<<endl;
return 0;
}
but with the set of inputs of 10.75(mealamonut), 17(tip%),5 (tax %).
the value am getting is 12.50
if I use
int totalSum = std::round(total);
its getting converted to 12
but my requirement is 13 .
How to acheive that ?
I cannot find any duplicate question to this if exists
please mention.

Answer 1

There are multiple ways to convert doubles to integers. You have multiple kinds of round. Check them here, which are std::round, std::lround, std::llround.
On the other hand, if what you want to do is not rounding, but elimination of fractions into one direction, then you have std::ceil that goes always higher, and std::floor, that always goes lower.
Remember to include <cmath>, not <math.h>. The latter is for C, not C++

Answer 2

You achieve your goal by using std::ceil() and std::floor() which is defined under cmath header file.

Answer 3

You are trying to always round up so you would need to use the ceil() function. Ceil is short for ceiling and there is also a floor function. Ceiling is up, floor is down, here is a c snippet to try out.
#include <stdio.h> /* printf */
#include <math.h> /* ceil */
int main ()
{
printf ( "ceil of 2.3 is %.1f\n", ceil(2.3) );
printf ( "ceil of 3.8 is %.1f\n", ceil(3.8) );
printf ( "ceil of -2.3 is %.1f\n", ceil(-2.3) );
printf ( "ceil of -3.8 is %.1f\n", ceil(-3.8) );
return 0;
}

Answer 4

for rounding to nearest integer math.h has nearbyint
printf ( "nearbyint (2.3) = %.1f\n", nearbyint(2.3) );
printf ( "nearbyint (3.8) = %.1f\n", nearbyint(3.8) );
Output:
nearbyint (2.3) = 2.0
nearbyint (3.8) = 4.0
Or if you want to break the default rounding behavior when .5
int totalSum= (total - floor(total) ) >= 0.5 ? ceil(total) : floor(total);

Answer 5

1) 10.75 + 17*10.75/100 + 5*10.75/100 = 13.115 ... how comes I can't get 12.50?
2) How do you know it's 12.50, how do you check value of result? (it may be only 12.4999..., so when it is formatted to two decimal places, it will become 12.50) Make sure you do check the actual real value (ideally in debugger or dump memory content in bytes and reconstruct the value by hand), not some string formatted intermediate.
3) this is not some production code, right? Amounts are not calculated with doubles in real financial software, as doubles are not accurate enough and you would run into all kind of hard problems with rounding to VAT, etc. If this is some real thing, you are not up to the task, ask for help some professional.
Answer:
std::round should normally do what you need. If it's ending as 12, then it's because the value is less than 12.5.
If rounded to two decimal places it shows as 12.50, you are hitting one of those "all kind of hard problems" of real financial software.
Then you should create your own round using string representation, like this example (not handling negative numbers and reinventing wheel probably):
#include <iostream>
#include <string>
/**
* Rounds floating point value in std::string type.
* Works only for positive values, and without "+" sign!
* ie. regexp ~\d*\.?\d*~ formatting.
* For malformed input the output is undefined (but should not crash).
**/
std::string financialRound(const std::string & amount) {
const size_t dotPos = amount.find_first_of('.');
if (std::string::npos == dotPos) return amount; // no decimals found
// copy whole part into temporary result
std::string result = (0 < dotPos) ? amount.substr(0, dotPos) : "0";
const size_t firstDecimalPos = dotPos + 1;
// see if there is 5 to 9 digit after dot
if (amount.size() <= firstDecimalPos) return result; // no char
const char firstDecimal = amount.at(firstDecimalPos);
if (firstDecimal < '5' || '9' < firstDecimal) return result; //not 5-9
// add 1 to the result
int patchPos = (int)result.size();
while (0 <= --patchPos) {
++result.at(patchPos);
if ('9'+1 == result.at(patchPos)) result.at(patchPos) = '0';
else break;
}
// check if additional 1 is required (add overflow)
if (-1 == patchPos) result.insert(result.begin(), '1');
return result;
}
void tryValue(const std::string & amount) {
printf("\"%s\" is rounded to \"%s\"\n", amount.c_str(), financialRound(amount).c_str());
}
int main()
{
printf("Trying normal values...\n");
tryValue("12.50");
tryValue("12.49");
tryValue(".49");
tryValue(".50");
tryValue("9.49");
tryValue("9.50");
printf("Missing decimals...\n");
tryValue("12");
tryValue("12.");
printf("Malformed...\n");
tryValue("");
tryValue("a.4");
tryValue("a.5");
tryValue("12..");
}
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