Why is result of double arithmetic wrong?

Question

I have the following, in C++:

    double clsTime::dblSSM(INT16U uint16Hrs, INT16U uint16Mins, INT16U uint16Secs, INT16U uint16Hunds) {
        double dblHrsAsSecs  = (double)(uint16Hrs % 24) * 3600.0
              ,dblMinsAsSecs = (double)(uint16Mins % 60) * 60.0
              ,dblSecs       = (double)(uint16Secs % 60)
              ,dblHundreths  = (double)(uint16Hunds % 100) / 100.0
              ,dblSSM = dblHrsAsSecs + dblMinsAsSecs + dblSecs + dblHundreths;     
        return dblSSM;                      
    }

When calling this function with:

    dblResult = dblSSM(16, 56, 17, 13);

I can see in the debugger that the results of the intermediate variables are:

    dblHrsAsSecs  = 57600
    dblMinsAsSecs = 3360
    dblSecs       = 17
    dblHundreths  = 0.13

However when combined into dblSSM the result is:

    60977.1

What I want is: 60977.13 which is the correct answer, why is the .03 being dropped and how can I get it included?

To convert back to a string for checking:

    void clsTime::ssmToString(double dblSSM, char* pszBuffer, char szFormat[]) {
      INT8U uint8Hours, uint8Minutes, uint8Seconds, uint8Hundreths;
      double dblDummy;

      assert(pszBuffer != NULL);
      long lngSSM = (long)dblSSM;  
      uint8Hours = (INT8U)(lngSSM / ((long)mscuint16SecsInHour));
      uint8Minutes = (INT8U)((lngSSM % ((long)mscuint16SecsInHour)) /
                                               (long)mscuint16SecsInMinute);
      uint8Seconds = (INT8U)(lngSSM % (long)mscuint16SecsInMinute);
      int8Hundreths = (INT8U)(modf(dblSSM, &dblDummy) * 100.0);  
      sprintf(pszBuffer, szFormat, uint8Hours, uint8Minutes
                         , uint8Seconds, uint8Hundreths);
    }

szFormat is an optional parameter, the default passed in is:

    "%02.2u:%02.2u:%02.2u.%02.2u"

How do you watch that value? Did you try to print the value for example with more digits: https://stackoverflow.com/a/554134/8051589? — Andre Kampling, Jul 13 '17 at 13:07
All the values are as displayed in the debugger, in watch window. If I use the same value to convert back to a string I can see the result is also less the .03 — SPlatten, Jul 13 '17 at 13:07
Can the debugger show more digits or just the one you see? Try to get more digits displayed. If it's not possible use std::cout as in the answer given above. — Andre Kampling, Jul 13 '17 at 13:09
@AndreKampling, as I said in my edit, the converted result back to a string shows the double is incorrect. — SPlatten, Jul 13 '17 at 13:10
@AndreKampling, I'll edit my post and include the other function to convert back. — SPlatten, Jul 13 '17 at 13:11
can you just print the results without debugger: `cout << dblSSM << all others too << endl`? — Serge, Jul 13 '17 at 13:15
The answer below from GediminasLe is absolutely right. But back to your code which format string are you using? For example use "%.3f" to get more digits. Further you tagged your code as C++ but it's more C like except of that your using a class. — Andre Kampling, Jul 13 '17 at 13:18
if you do this line: double d = 60977.13; what's the value that your debugger shows: 60977.13 or 60977.129999999997 ? — João Augusto, Jul 13 '17 at 13:59
@JoãoAugusto, the debugger shows the correct result, I've event created an expression in the debugger which demonstrates the problem isn't the debugger. — SPlatten, Jul 13 '17 at 14:00
If It shows 60977.13, then your debugger is rounding the value for display purposes. — João Augusto, Jul 13 '17 at 14:03

GediminasLe · Accepted Answer · 2017-07-13T13:48:42.540

2

I think it's not the result but your printing problem.
I have written this program:

#include <iostream> // cout
#include <iomanip> // fixed, setprecision

using namespace std;

double dblSSM(__uint16_t uint16Hrs, __uint16_t uint16Mins, __uint16_t uint16Secs, __uint16_t uint16Hunds) {
        double dblHrsAsSecs  = (double)(uint16Hrs % 24) * 3600.0
              ,dblMinsAsSecs = (double)(uint16Mins % 60) * 60.0
              ,dblSecs       = (double)(uint16Secs % 60)
              ,dblHundreths  = (double)(uint16Hunds % 100) / 100.0
              ,dblSSM = dblHrsAsSecs + dblMinsAsSecs + dblSecs + dblHundreths;     
        return dblSSM;                      
    }

int main () {

    cout << dblSSM(16, 56, 17, 13) << endl; // prints 60977.1
    cout << fixed << setprecision(8)/*output 8 in the decimal part */ << dblSSM(16, 56, 17, 13) << endl; // prints 60977.13000000

    return 0;
}

I think it works as it should.
If you want to print double you should use setprecision and fixed.

EDIT:

cout << fixed << setprecision(12) << ( modf(dblSSM, &dblDummy) * 100.0);

Prints 12.999999999738 so (__uint16_t)(modf(dblSSM, &dblDummy) * 100.0) gives 12. Hacky way to fix this is to use round function, but I suggest you not use doubles unless you really have.

edited Jul 13 '17 at 13:48

answered Jul 13 '17 at 13:13

GediminasLe

44
5

1

The fact this answer has both [`bits/stdc++.h`](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) and [`using namesapce std;`](https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice) makes it a horrible suggestion to any future reader. – StoryTeller - Unslander Monica Jul 13 '17 at 13:19
1

Retracted DV. Still, the using directive makes me cringe. – StoryTeller - Unslander Monica Jul 13 '17 at 13:25
But its not a display problem, why is dblSSM wrong? How do you fix that? – SPlatten Jul 13 '17 at 13:49
@SPlatten function `clsTime::dblSSM` works correctly, where is problem in `clsTime::ssmToString`. – GediminasLe Jul 13 '17 at 13:53
@GediminasLe, so what is the solution to getting dblSSM correct, I'm not interested in correcting the display, just the correct result. dblSSM isn't correct because it returns a value without the .03 – SPlatten Jul 13 '17 at 13:58
@SPlatten `dblSSM` is correct. `dblSSM(16, 56, 17, 13)` returns `60977.13` – GediminasLe Jul 13 '17 at 14:00
Not with the compiler I'm using which is the now rather old Borland C++ 5.2, I don't have a choice on the compiler as its for an old embedded system. – SPlatten Jul 13 '17 at 14:36
@Splatten if you do: double d = 60977.13; unsigned int b = d * 100; what's the value of b on your debugger ? – João Augusto Jul 13 '17 at 14:46
@JoãoAugusto, same, I assigned value of dblTest = 60977.13 and the value actually displayed in the debugger is 60977.1, so is this an optimization? – SPlatten Jul 13 '17 at 14:50
@Splatten, and what's the value of variable b (unsigned int b = dblTest * 100) ? If it's 6097713 then youre debugger is rounding values for display purposes. – João Augusto Jul 13 '17 at 14:58
@Splatten, that's because you declared b as a short, please use a uint32 – João Augusto Jul 13 '17 at 15:09
Its been a long day :( – SPlatten Jul 13 '17 at 15:10
@JoãoAugusto, now the value of b is 6097712 – SPlatten Jul 13 '17 at 15:12
@Splatten, so as you can see it's the debugger that is rounding for display purposes. – João Augusto Jul 13 '17 at 15:25

Why is result of double arithmetic wrong?

1 Answers1