C++ : Issue with division in loop [duplicate] - c++

Question

This question already has answers here:
Why does dividing two int not yield the right value when assigned to double?
(10 answers)
Closed 6 years ago.
i am learning c++ , student of first semester,
I am trying to divide 2 arrays in loop and store its result in 3rd array that's not working.All other operations work except division following is my code
#include<iostream>
using namespace std;
int main(){
int arr[6]={10,20,30,40,50,06};
int arr2[6]={10,20,20,30,40,50};
int input;
cout<<"please enter 1 for addition.... "<<endl;
cout<<"please enter 2 for subtraction.... "<<endl;
cout<<"please enter 3 for multiplication.... "<<endl;
cout<<"please enter 4 for division.... "<<endl;
cout<<"please enter 5 for %age...."<<endl;
cin>>input;
float arr3[6];
int i,j,k,l,m;
switch(input){
case 1:
for(i=0;i<6;i++){
arr3[i]=arr[i]+arr2[i] ;
cout<<arr3[i]<<endl;
}
break;
case 2:
for(j=0;j<6;j++){
arr3[j]=arr[j]-arr2[j];
cout<<arr3[j]<<endl;
}
break;
case 3:
for(k=0;k<6;k++){
arr3[k]=arr[k]*arr2[k];
cout<<arr3[k]<<endl;
}
break;
case 4:
for(l=0;l<6;l++){
arr3[l]=arr[l]/arr2[l];
cout<<arr3[l]<<endl;
}
break;
case 5:
for(m=0;m<6;m++){
arr3[m]=arr[m]/(arr[m]+arr2[m]);
cout<<arr3[m]<<endl;
}
break;
}
}
for division output is :
1
1
1
1
1
0
i tired to set its data type as dobule and float both but its not working need your help please.

Answer 1

You need to cast the operands before you divide, or else it will use integer division, which has its decimal shaved off. Using c-style casts, it would look like:
arr3[l]= (double)arr[l] / (double)arr2[l];
Note, C++ has its own casting syntax, but it's been awhile since I've used them, and can't remember them exactly. I think it would look like:
arr3[l] = static_cast<double>(arr[l])
/ static_cast<double>(arr2[l]);