Why does using my print method with std::cout result in an error?

Question

#include <iostream>

using namespace std;

class Fam
{
    public:
        char you, urmom, urdad;
        void addPerson(char y, char m, char f)
        {
            you = y;
            urmom = m;
            urdad = f;
        }
};

class Tree: public Fam
{
    public:
        void showFamtree()
        {
            cout<< "Name: " << you << endl;
            cout<< "Mother's name: " << urmom <<endl;
            cout<< "Father's name: " << urdad <<endl;
        }
};

int main(void)
{
    Tree tree;

    char a,b,c;
    cin >> a;
    cin >> b;
    cin >> c;

    tree.addPerson(a,b,c);

    cout<< "Family tree: " << tree.showFamtree() <<endl;

    return 0;    
}

I wanted to print the family tree with the person's name, mother's name, father's name but when I compile it, I get the following error:

invalid operands to binary expression (basic_ostream<char, std::char_traits<char> > and void)

Please don't tag unrelated languages, this is certainly not C. — Baum mit Augen, May 25 '16 at 01:32
I edited your post to give it a title that carries some information and I included some error message for your problem. (The latter was generated with clang, you may change that to the error message your compiler yields if you want to.) For the future, please make sure your titles describe the problem and your posts include relevant error messages. Have fun on this site. — Baum mit Augen, May 25 '16 at 01:44
Thanks, everyone. All apologies if my question didn't make that much sense to you. I'm just a Computer Science student trying to save my grades this semester so I can finally shift to my desired major. — ree, May 25 '16 at 04:47

score 2 · Answer 1 · answered May 25 '16 at 01:33

2

tree.showFamtree() returns nothing (i.e. void), it doesn't make any sense to try to pass it to std::cout. You might change

cout<< "Family tree: " << tree.showFamtree() <<endl;

to

cout << "Family tree: " << endl;
tree.showFamtree();

answered May 25 '16 at 01:33

songyuanyao

169,198
16
310
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BLUEPIXY · Answer 2 · 2016-05-25T02:00:18.027

1

If you define operator << like this

ostream& operator << ( ostream & ostr , const Tree & t ){
    ostr << "Name:" << t.you << endl
        << "Mother's name:" << t.urmom << endl
        << "Father's name:" << t.urdad << endl;
    return ostr;
}

then you can use

cout<< "Family tree: " << tree <<endl;

This is known as operator overloading in C++.

edited May 25 '16 at 02:00

answered May 25 '16 at 01:53

BLUEPIXY

39,699
7
33
70

Why `static` function? – songyuanyao May 25 '16 at 01:58
Why the `static`? € Too slow. – Baum mit Augen May 25 '16 at 01:58
There is no meaning much. I remove it. – BLUEPIXY May 25 '16 at 01:59
@BLUEPIXY You're welcome. For a complete answer, you should explain what's wrong with the current approach IMO. (Already upvoted though as I guess you explained what OP is trying to do.) – Baum mit Augen May 25 '16 at 02:21
it was omitted because it was already described @songyuanyao is the cause when I try to answer. – BLUEPIXY May 25 '16 at 02:36

2785528 · Answer 3 · 2016-05-25T03:40:20.490

To use something similar to this:

void showFamtree()
{
    cout<< "Name: " << you << endl;
    cout<< "Mother's name: " << urmom <<endl;
    cout<< "Father's name: " << urdad <<endl;
}

with:

cout << "Family tree: " << tree.showFamtree() << endl;

One C++ approach would be to use std::stringstream, as in:

std::string showFamtree()
{
    std::stringstream ss;
    ss << "Name: " << you << endl;
    ss << "Mother's name: " << urmom <<endl;
    ss << "Father's name: " << urdad <<endl;
    return (ss.str());
}

I also often add a label. So consider using

std::string showFamtree(std::string label)
{
    std::stringstream ss;
    ss << label;
    ss << "Name: " << you << endl;
    ss << "Mother's name: " << urmom <<endl;
    ss << "Father's name: " << urdad <<endl;
    return (ss.str());
}

and change invocation to

cout << tree.showFamtree("Family tree: ") << endl;

Note - Perhaps the label should be on its own line, for consistent white space on the left of the 'tree'.

Why does using my print method with std::cout result in an error?

3 Answers3