c++ - 为什么将我的打印方法与 std::cout 一起使用会导致错误？

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;    
}

我想用人的名字，母亲的名字，父亲的名字打印家谱，但是当我编译它时，出现以下错误：

basic_ostream<char, std::char_traits<char> >二进制表达式 ( and void)的无效操作数

Answer 1

tree.showFamtree()什么都不返回（即void），尝试将它传递给 . 没有任何意义std::cout。你可能会改变

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

至

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

Answer 2

如果你operator <<这样定义

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;
}

那么你可以使用

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

这在 C++ 中称为运算符重载。

Answer 3

要使用类似的东西：

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

和：

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

一种 C++ 方法是使用 std::stringstream，如：

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());
}

---

我也经常加标签。所以考虑使用

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());
}

并将调用更改为

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

---

注意 - 也许标签应该在自己的行上，以便在“树”左侧保持一致的空白。