CODEWITHSUNDEEP
cunsigned-integer
aliants
aliants

Reputation: 77

Wrong result in operation with unsigned integers

I have a problem with an arithmetic operation with unsigned integer variables.

All the variables are defined as uint32_t. This is the arithmetic operation:

batt += (uint32_t) ((((charg - discharg) * (time_now - time_old)) / 1000) + 0.5);

The values before the operation are:

batt = 8999824
charg = 21
discharg = 1500
time_now = 181
time_old = 132

The problem is that the result after the operation is

batt = 13294718

instead of

batt = 8999752

What's the reason?

Thanks in advance.

Upvotes: 1

Views: 440

Answers (2)

chux
chux

Reputation: 154169

You have 2 problems.

  1. charg < discharg as so creates a wrap-around answer of 4294965817 for charg - discharg. See below as to why you ended up with 13294718.

  2. Do the biasing (+ 0.5) before the /1000, else the integer division will all ready have tossed the fractional part.

Recommended fix 1: insure charg >= discharg.

OR

Recommended fix 1: change charg, discharg, time_now, time_old and maybe batt to int32_t.

Recommended fix 2: change your rounding to batt += (uint32_t) ((Product / 1000.0) + 0.5);

OR

Recommended fix 2: change your rounding to batt += (Product + 500*sign(Product))/1000;

Apparent errant code - step by step.

uint32_t batt = 8999824;
uint32_t charg = 21;
uint32_t discharg = 1500;
uint32_t time_now = 181;
uint32_t time_old = 132;
// batt += (uint32_t) ((((charg - discharg) * (time_now - time_old)) / 1000) + 0.5);

// The big problem occurs right away.
// Since charg is less than discharg, and unsigned arithmetic "wrap around", 
// you get (21 - 1500) + 2**32 = (21 - 1500) + 4294967296 = 4294965817
uint32_t d1 = charg - discharg;
uint32_t d2 = time_now - time_old; // 49
// The product of d1 and d2 will overflow and the result is mod 4294967296
// (49 * 4294965817) = 210453325033
// 210453325033 mod 4294967296 = 4294894825
uint32_t p1 = d1 * d2;
uint32_t q1 = p1/1000;  // 4294894825/1000 = 4294894.825.  round to 0 --> 4294894
double s1 = q1 + 0.5;  //  4294894 + 0.5 --> 4294894.5;
uint32_t u1 = (uint32_t) s1; // 4294894.5 round to 0 --> 4294894
batt += u1; // 8999824 + 4294894 --> 13294718

Upvotes: 1

Alex
Alex

Reputation: 10136

The result of the charg - discharg is negative, thus all expression is a negative, that is a pretty big unsigned.

Upvotes: 1

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