CODEWITHSUNDEEP
c++vectoroperator-overloadingoperators
Coder95
Coder95

Reputation: 131

C++ operator overloading << with vector

Hey guys I am new to C++ and I have a problem with this operator: (Also new in stackoverflow)

This is my class TestList:

class TestList{
public:
    TestList() : listItems(10), position(0){};
    TestList(int k) : listItems(k), position(0){};
    int listItems;
    int position;
    std::vector<int> arr;
};


//my current operator is: What should be changed?
ostream& operator <<(ostream&, const TestList& tlist, int input){
    os << tlist.arr.push_back(input);
    return os;
}
//

int main() {
testList testlist(5);
 testlist << 1 << 2 << 3; //how should I overload the operator to add these number to testlist.arr ?
 return 0;
}

I hope someone could help me or can give me any tips? :)

Upvotes: 1

Views: 88

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 310920

I think you mean the following

TestList & operator <<( TestList &tlist , int input )
{
    tlist.arr.push_back( input );
    return tlist;
}

Here is a demonstrative program

#include <iostream>
#include <vector>

class TestList{
public:
    TestList() : listItems(10), position(0){};
    TestList(int k) : listItems(k), position(0){};
    int listItems;
    int position;
    std::vector<int> arr;
};

TestList & operator <<( TestList &tlist , int input )
{
    tlist.arr.push_back( input );
    return tlist;
}

std::ostream & operator <<( std::ostream &os, const TestList &tlist )
{
    for ( const auto &item : tlist.arr )
    {
        std::cout << item << ' ';
    }

    return os;
}

int main() 
{
    TestList testlist(5);

    testlist << 1 << 2 << 3;

    std::cout << testlist << '\n';

    return 0;
}

The program output is

1 2 3

You can even write instead of these two statements

testlist << 1 << 2 << 3;

std::cout << testlist << '\n';

only one statement

std::cout << ( testlist << 1 << 2 << 3 ) << '\n';

Pay attention to that there is a typo in your declaration

testList testlist(5);

There should be

TestList testlist(5);

Upvotes: 0

n314159
n314159

Reputation: 5075

The other answers are absolutely correct, I just want to say something general on operator<<. It always has the signature T operator<<(U, V), since it is always a binary operator, so it has to have exactly two arguments. Since the chain 

a << b << c;

is evaluated as

(a << b) << c;
// That calls
operator<<(operator<<(a, b), c);

the types T and U should normally be the same, or at least compatible.

Furthermore, it is possible but very weird to assign the result of operator<< to something (like result = (a << b))). A good rule of thumb is "My code should not be weird". Therefore the type T should mostly be a reference (so X&) since otherwise it would only be a temporary copy that is unused. And that is pretty useless most of the time.

So in 90% of all cases, your operator<< should have the signature T& operator<<(T&, V);

Upvotes: 1

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