Invalid operand of types 'long long int' to binary 'operator%'

Question

i am in C++ and I get this error:

bool comprovarCodi(long long num, int DC){
bool codi_correcte;
int i=0, suma_senars=0, suma_parells=0, suma_total=0, desena_superior, DC_calculat, cont=0;
while(num!=0){
    num=num/10;
    cont++;
    i++;    
}
if(cont==12){
    for(int j=1; j<12; j=j+2){
        suma_senars=suma_senars+num%pow(10,j);

I don't know why, I believe "num" is an integer so I can use the operator "%".

Somebody knows why it fails?

Thank you

`pow` returns `double`, you cannot apply `%` to `long long` and `double`. — AlexD, Dec 30 '15 at 20:44
You need to read the _whole_ error message, not just half of it. — Lightness Races in Orbit, Dec 30 '15 at 20:45
(pow also exists for float and long double, but yes, no integer return type) — deviantfan, Dec 30 '15 at 20:46
The problema is that I need to get every digit of a number of 12 digits. That is why I use the long long. Do you know another method to do it? — user3894819, Dec 30 '15 at 20:47
@user3894819 As said already by multiple people, the problem is not `long long`, but `double` (`pow`) — deviantfan, Dec 30 '15 at 20:48

score 1 · Accepted Answer · answered Dec 30 '15 at 20:56

Don't use pow for this sort of thing.

long long pow_ten = 10;
for(int j=1; j<12; j=j+2)
{
   suma_senars=suma_senars+num%pow_ten;
   pow_ten *= 100;
}

Not only will this be faster, it will also calculate correctly, rather than pow which may well use something like exp(log(x) * y) to calculate x ** y - and thus not always come up with precisely the number you wanted - particularly if you cast it back to integer.

score 0 · Answer 2 · answered Dec 30 '15 at 20:48

0

you have to cast to int first
suma_senars = suma_senars + num % (int)pow(10,j);
better: suma_senars += num % (int)pow(10,j);
clearer: suma_senars += num % ( (int)pow(10,j) );

answered Dec 30 '15 at 20:48

I3ck

433
3
10

Actually, I think your clearer line is less clear than your better line -- extra redundant parenthesis generally make things harder to read IMO. – Chris Dodd Dec 30 '15 at 20:53
2

It is a bad idea, since you can easily get unexpected result due to cutting of the floating part. `(int)9.9999` is `9`, not `10`. In general, as **@Mats** said, do not use `pow` in such scenarios. – AlexD Dec 30 '15 at 20:58

Invalid operand of types 'long long int' to binary 'operator%'

2 Answers2