Arithmetic expression is 0 instead of 1

Question

#include <iostream>
using namespace std;
int main() {
  int a,b,c;
  cin >> a >> b >> c;
  cout << a/b*c;
  return 0;
}

For a=5 b=10 c=2 the output gives 0 which is obviously wrong. As far as I understood << / * are binary operators / and * "bond" stronger than << and for / and * it is calculated from the left to right so first 5 is devided by 10, then the result (0.5) is multiplicated with 2 which is 1 and that is delivered by << to cout.

So can anyone give me an explanation for that result (0)?

you should cast your division to double to get your expected result — Rodrigo López, May 01 '15 at 20:28
If you want to keep it integer arithmetic, you could rearrange your expression: `c*a/b`. Of course, you'll get truncation when `c*a` is not divisible by `b`. — Fred Larson, May 01 '15 at 20:32
@Chromo sapiens: Not exactly. There never was a `0.5` in your operation. When you divide integers, you immediately get a whole quotient and a reminder. `5 / 10` gives `0` quotient. `5 % 10` gives `5` as a remainder. — AnT stands with Russia, May 01 '15 at 20:32

score 2 · Answer 1 · answered May 01 '15 at 20:26

2

This is the integer division issue: if a==5 and b==10, then a/b==0.

See the standard, Multiplicative operators [expr.mul]

For integral operands the / operator yields the algebraic quotient with any fractional part discarded

answered May 01 '15 at 20:26

AlexD

32,156
3
71
65

score 1 · Answer 2 · answered May 01 '15 at 20:32

This is the result of integer rounding. When integer division is performed any fractional remainder is simply cut off. So 2 / 3 == 0.

To keep the results as accurate convert use doubles or convert the variables to doubles in the expression. So something like this

static_cast<int>(static_cast<double>(a) / static_cast<double>(b) * static_cast<double>(c))

will result in the correct value.

score 0 · Answer 3 · answered May 01 '15 at 20:27

No its not wrong

Since all variables are integers.

5/10 is 0 in the integer world and that multiplied with 2 is still 0

9/10 is also 0 but 11/10 is 1

if you want something useful from this declare the variables as double instead or IOW:

int main() {
  double a,b,c;
  cin >> a >> b >> c;
  cout << a/b*c;
  return 0;
}

Arithmetic expression is 0 instead of 1

3 Answers3