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c++ sum of even and odd numbers

Thoughtless Toucan on Dec 30, 2020

//given the fact that it takes n of number
void sumEvenAndOdd(){
int n=0 , evenSum=0 , oddSum=0;
     int numbers[Size];
     cout<<"enters the number(N) of integers\n";
     cin>>Size;

     cout<<"Enter "<<Size<<" number/s:\n";
     for(int i =0 ;i< Size ; ++i){
        cin>>numbers[i];
     }
     for(int i =0 ;i<Size ; ++i){
        if(numbers[i]%2 ==0){
            evenSum += numbers[i];
        }
        else{
            oddSum += numbers[i];
        }
     }
     cout<<"the sum of even numbers is : "<<evenSum <<"\nthe sum of odd numbers is :"<<oddSum<<endl;
}

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c++ sum of even and odd numbers Given the following declarations below. Write a loop to read a list of numbers from the keyboard terminated by -999 and store the even numbers (skip over the odd numbers) in the vector v. Read in three numbers, and calculate the sum. Output the sum as an integer. in c visual studio return odd numbers c++ return odd numbers c++ for loop sum of 2 numbers in cpp c++ sum up numbers sum of 2 numbers in cpp function Sum of two large numbers in C++ c++ sum up numbers c++ sum up numbers sum of 2 numbers in cpp Sum of two large numbers in C++ Sum of two large numbers in C++ sum of n natural numbers in c sum of 2 numbers in cpp function kadane algorithm with negative numbers included as sum even and odd in c++ c++ print the amount of odd integer between n and m c++ print the amount of odd integer between n and m how to check if the number is even or odd using bitwise operator C++ sum a vector of digits sum elements in vector c++ sum of number sum array c++ sum array c++ sum of vector c++ sum of stack c++ sum of vector c++ sum of vector c++

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"c++ sum of even and odd numbers" Code Answer's