如何从C ++中的函数返回向量？

Question

vector<double> function(vector<double> &X){
    vector<double> data2;
    for (int i = 0; i < X.size(); i++){
        data2.push_back(2*X[i]);
    }
    return data2;
}
int main(){
    vector<double> data;
    for (int i = 1; i <= 10; i++){
        data.push_back(i);
    }
    for (int i = 0; i < data.size(); i++){
        cout << function(data) << endl;
    }
return 0;
}

Basically, for a given artificially created vector of "data" for data[i] = i, I want a function such that it multiplies each element of the vector by 2, then print the result. However, I can't seem to understand what I did wrong.

Answer 1

The function returns a std::vector which is a kind of container. And we can't use std::cout to print the elements of std::vector .
We should go into the container to get elements and print them.
Like this:

data2 = function(data);
for (int i = 0; i < data2.size(); i++)
{
        cout << data2[i]<< endl;
}

Answer 2

First of all, to simplify things, let's write a simple function that prints out the contents of a vector. Now, to see what a vector looks like we can just call the function and see what it looks like.

template <typename T>
void printVector(const vector<T> &input){
    unsigned sz = input.size();
    cout<<"========"<<endl;
    for(unsigned i=0; i<sz; i++){
        cout<<input[i]<<endl;
    }
    cout<<"========"<<endl;
}

Remember, if a function takes an object of a generics class (such as std::vector< T >), you need to specify that it is a template function.

Now, back to your question. I am assuming that you do not want change the values of data itself, because you declared data2 .

Then you have two options (if I understand the problem you are trying to solve).

The first option is to write a function that returns a vector.

#include <iostream>
#include <vector>

using namespace std;

template <typename T>
vector<T> doublingFunc(const vector<T> &input){
    vector<T> output;
    /* it is good practice
       to catch the size of the vector
       once so that you aren't calling
       vector::size() each go through
       of the loop, but it is no big deal
    */
    unsigned sz = input.size();
    for(unsigned i=0; i<sz; i++){
        output.push_back(2*input[i]);
    }
    return output;
}

template <typename T>
void printVector(const vector<T> &input){
    unsigned sz = input.size();
    cout<<"========"<<endl;
    for(unsigned i=0; i<sz; i++){
        cout<<input[i]<<endl;
    }
    cout<<"========"<<endl;
}

int main(){
    vector<double> data;

    for(int i=1; i<10; i++){
        data.push_back((double) i);
       //technically, the cast in unnecessary
    }

    printVector(data);

    vector<double> data2 = doublingFunc(data);

    printVector(data2);

    return 0;
}

The second option is to write a function that dynamically allocates a new vector and then return a pointer to it.

#include <iostream>
#include <vector>

using namespace std;

template <typename T>
vector<T>* doublingFunc(const vector<T> &input){
    vector<T>* output = new vector<T>();
    /* it is good practice
       to catch the size of the vector
       once so that you aren't calling
       vector::size() each go through
       of the loop, but it is no big deal
    */
    unsigned sz = input.size();
    for(unsigned i=0; i<sz; i++){
        output->push_back(2*input[i]);
    }
    return output;
}

template <typename T>
void printVector(const vector<T> &input){
    unsigned sz = input.size();
    cout<<"========"<<endl;
    for(unsigned i=0; i<sz; i++){
        cout<<input[i]<<endl;
    }
    cout<<"========"<<endl;
}

int main(){
    vector<double> data;

    for(int i=1; i<10; i++){
        data.push_back((double) i);
    }

    printVector(data);

    vector<double>* data2 = doublingFunc(data);

    /*this function takes a reference to a vector,
     so we need to dereference the pointer
   */
    printVector(*data2);

    //remember to delete dynamically allocated variables
    delete data2;
    return 0;
}

Of course, if you just want to print out twice the value of all entries in a vector, you can just use the function:

template <typename T>
void printDouble(const vector<T> &input){
    unsigned sz = input.size();
    cout<<"========"<<endl;
    for(unsigned i=0; i<sz; i++){
        cout<<2 * input[i]<<endl;
    }
    cout<<"========"<<endl;
}

This might be a lot of new stuff for you, as you seem new to c++. I suggest reading through this website.

Answer 3

For std::cout to print an std::vector, you have to overload the stream insertion operator (<<) for it.

eg

std::ostream& operator << ( std::ostream& os, const std::vector<double>& v )
{
    os << "{ ";
    for ( const auto& i : v )
    {
        os << i << ' ';
    }
    os << "}";
    return os;
}

For printing the contents of a container, you can use C++11 range-for loop as used in the above example.

You can also use std::copy algorithm for this overload or use it directly like this:

std::ostream& operator << ( std::ostream& os, const std::vector<double>& v )
{
    os << "{ ";
    std::copy( begin(v), end(v), std::ostream_iterator<double>( os, " " ) );
    os << "}";
    return os;
}

In your code, you can simply use function to operate on the values of the vector and then print it like this:

void multiplyBy2( vector<double>& v )
{
    for ( auto& i : v )
    {
        i *= 2;
    }
}

int main( void )
{
    vector<double> v;
    // ...
    multiplyBy2( v );
    std::cout << v << std::endl;

    return 0;
}