0

my expected result is average=73.5 ,i have set the type of average as double but it result 73 what's the problem?

  #include <iostream>
  using namespace std;

  int main(){
int x=0;
int total=0;
double average=0;
int counter=0;

cout<<"Question 1"<<endl<<"Enter integer(-100 to end);";
cin>>x;
 if (x!=-100)
 {
     for(;x!=-100;counter++)
     {
         total=total+x;
         cin>>x;
     }

      average = total/counter;
 }
 cout<<"The average is:"<<average<<endl;


 return 0 ;

}

Griwes
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alexllooyyoo
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  • since `total` and `counter` are both `int`, the result of that calculation will be `int`. There are many ways to solve it - `(1.0*total)/counter` is one. – Floris Nov 03 '13 at 15:15

2 Answers2

2

You're doing integer calculations. Cast one of the integers to double:

average = ((double)total)/counter;
Maroun
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Yochai Timmer
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  • Just for curiosity, would static_cast be preferable over (double)? – lolando Nov 03 '13 at 15:18
  • Just a syntactic difference. It's exactly the same in this case. – Yochai Timmer Nov 03 '13 at 15:23
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    @lolando: I would certainly prefer `static_cast(total)`: although in this case it does exactly the same, it would catch problems when it does _not_ the same, e.g., when the C-style cast would result in a `reinterpret_cast<...>()` or a `const_cast<...>()`. – Dietmar Kühl Nov 03 '13 at 15:25
1

Integer operations yield integers as result. In C and C++ they never yield floating point results. You need to involve a floating point value in the computation, e.g.

average = (1.0 * total) / counter;
Dietmar Kühl
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