通过引用语法传递的C ++向量

Question

I am having issues passing the the return values of my reference into the main function.The code is supposed to return both the the amount of even and odd numbers the user enters. I think my syntax in the pass are wrong.

  using namespace std;

int numOfOddEven(vector<int>);
int main ()
{
int numValues, odd;
vector<int> values;
cout << "Enter number of values: ";
cin >> numValues;

for(int count = 0; count < numValues; count++)
{
    int tempValue;
    cout << "Enter value: ";
    cin >> tempValue;
    values.push_back(tempValue);
}

 cout <<"Odd numbers: " << numOfOddEven(odd);
}cout <<"Even numbers: " << numOfOddEven(even);


int numOfOddEven(vector<int> vect)
{
int odd = 0;
int even = 0;

for (int i = 0; i < vect.size(); i++)
    if (i/2 == 1 )
        ++even;
    else
        ++odd;

return odd;
return even;
}

Answer 1

Handful of things I see wrong with this code

1. You cannot return two elements like you are trying
2. You are not passing by reference anywhere
3. What you are passing to numofOddEven is not what the function is expecting(an int vs a vector )

Learn how functions work, what pass by reference actually means, what return does, and what taking a modulus of a number means in c++. Then try approaching this again.

Answer 2

You're calling with wrong parameters and using wrong logic

 odd =numOfOddEven(values); //Use values
 cout <<"Odd numbers: " << odd;
 cout <<"Even numbers: " << values.size()- odd; //Get even count using this

In numOfOddEven just return the odd count

Fix Logic : -

int numOfOddEven(vector<int> vect)
{
int odd = 0;
//int even = 0;

for (int i = 0; i < vect.size(); i++)
    if (vect[i]%2 != 0 )
        ++odd;

return odd;
 //return even;
}

Another approach would be to use std::count_if

int numodd = std::count_if(values.begin(), values.end(), 
                      [](int i) {return i % 2 != 0;});
int numeven = values.size() - numodd ;

Answer 3

int numOfOddEven(vector<int> & vect)
{
int odd = 0;
int even = 0;

for (int i = 0; i < vect.size(); i++)
    if (vect[i]%2 == 1 )
        ++even;
    else
        ++odd;

return odd;

}


int main ()
{
int numValues, odd;
vector<int> values;
cout << "Enter number of values: ";
cin >> numValues;

for(int count = 0; count < numValues; count++)
{
    int tempValue;
    cout << "Enter value: ";
    cin >> tempValue;
    values.push_back(tempValue);
}

 cout <<"Odd numbers: " << numOfOddEven(values);
 cout <<"Even numbers: " << numValues - numOfOddEven(values);
 cin.get();
 cin.get();
 return 0;
}

Answer 4

References are defined in the function declaration/signature. For instance:

void fct( int &ref1, int &ref2)
{
   ref1 = 1;
   ref2 = 2;
}

int ref1, ref2;
fct(ref1, ref2);

No need for any returns. The compiler, when it sees the &, it considers it as a pointer but in code, you consider it as a variable.

Answer 5

To address the question in your subject:

int numOfOddEven(vector<int> vect) { }

Add a & before vect :

int numOfOddEven(vector<int> & vect) { }

Then vect will be passed by reference instead of copied. In regards to your return values, pass them in as references also and then declare the function of type void:

void numOfOddEven(vector<int> & vect, int & countodd, int & counteven) { }

Then just modify those variables in the function and don't return anything.

Answer 6

First of all, you are not passing anything by reference in your existing code. If you want to pass the vector by reference, you need to declare your function like this:

int OddCount(const std::vector<int>& v)
// ................................^ That denotes a reference
{
    return std::count_if(v.begin(), v.end(), [](int i)
    {
        return i % 2;
    });
}

int EvenCount(const std::vector<int>& v)
{
    return std::count_if(v.begin(), v.end(), [](int i)
    {
        return !(i % 2);
    });
}

NOTE : Your logic for determining an odd/even is fixed in both of the above functions. Your method is not correct (unless you think 2 is the only even number).

Second, you never declared an even vector (and there is no need to, nor is there a need to declare an odd vector). So you should modify your output statements:

cout << "Odd Numbers:  " << OddCount(values) << std::endl;
cout << "Even Numbers: " << EvenCount(values) << std::endl;

If you want both values to be returned from a single function call, there are a few ways to do it. The "simplest" way would be to return a std::pair<int, int> :

std::pair<int, int> CountOddsAndEvens(const std::vector<int>& v)
{
    int evens = std::count_if(v.begin(), v.end(), [](int i)
        {
            return !(i % 2);
        });

    int odds = v.size() - evens;

    return std::make_pair(evens, odds);
}

Alternatively, you could pass them in as output parameters:

void CountOddsAndEvens(const std::vector<int>& v, int& evens, int& odds)
{
    evens = std::count_if(v.begin(), v.end(), [](int i)
        {
            return !(i % 2);
        });

    odds = v.size() - evens;
}

Both of which would require you to make changes to your current output calls:

std::pair<int, int> results = CountOddsAndEvens(values);
std::cout << "Odds = " << results.second << ", Evens = " << results.first << std::endl;

Or

int evens, odds;
CountOddsAndEvens(values, evens, odds);
std::cout << "Odds = " << odds << ", Evens = " << evens << std::endl;

Answer 7

The first place to look is the declaration:

int numOfOddEven(vector<int>);

This returns one int , not two. You have to change something if you want to return two (independent) pieces of information. If you want to return two ints, you could write a function with this type:

pair<int,int> numOfOddEven(vector<int>);

Then, at the end of the function, you can return with return make_pair(num_odds, num_evens) . Then, you would need also to do something like this to actually accept the returned values:

tie(odd, even) = numOfOddEven(values);

---

But , that's probably too complex for a beginner. You want to find another, roundabout, way to "return" two numbers from one function call.

void numOfOddEven(vector<int>, int&, int&);

Note the return type is not void . This function doesn't really return anything. But you can pass in your two variables odd and even by reference to this. When you say "passing by reference", this is probably what you mean.

... more code required ... [ community-wiki :-) ]

---

But, again! It's obvious that every number is odd or even. Therefore, it is sufficient to just return one number, the odd number. Then you can calculate (within main ) that the number of even numbers is simply the total size of the vector minus the number of odd numbers.