C++输出运算符重载

发布于 2024-11-10 05:14:58 字数 998 浏览 3 评论 0原文

我一直在做这项学校作业。赋值告诉我们创建一个重载其输出运算符 ( << ) 的对象。这是我的代码：

#include <ostream>
using namespace std;

template <class T>
class CustomObject {

        string print() {
            string text = "";
            for (int i = 0; i < num_items(); i++) {
                text += queue[i];
                text += " | \n";
            }
            return text;
        }

        friend std::ostream& operator <<(std::ostream &output, CustomObject &q) {
            output << "" << q.print();
            return output;
        }
}

所以我像这样实例化这个对象：

CustomObject<int> co();

并调用它的输出方法：

std::cout << co();

这将不可避免地调用 print 方法，并将字符串返回到默认输出流。

但是，我的控制台/调试器中没有可见的输出。

我在这里缺少什么？

PS，这不是完整的类，它是通用的，因为这里没有必要显示其他一些方法和功能。

PPS num_items() 和队列变量是所说的其余部分，这个类是一个 PriorityQueue 对象。因此，queue 是指定类型的数组（因此是通用声明），num_items() 仅返回数组的计数。

原文

I've been working on this school assignment. The assignment told us to make an object which had it's output operator ( << ) overloaded.
Here's my code:

#include <ostream>
using namespace std;

template <class T>
class CustomObject {

        string print() {
            string text = "";
            for (int i = 0; i < num_items(); i++) {
                text += queue[i];
                text += " | \n";
            }
            return text;
        }

        friend std::ostream& operator <<(std::ostream &output, CustomObject &q) {
            output << "" << q.print();
            return output;
        }
}

So I instantiate this object like this:

CustomObject<int> co();

and call its output method:

std::cout << co();

Which would inevitably call the print method, and return the string to the default output stream.

But, there's no visible output in my console/debugger.

What am I missing here?

PS this is not the complete class, it's generic because of several other methods and functionality that is not necessary to be shown here.

PPS the num_items() and queue variables are part of said rest, this class is a PriorityQueue object. So, queue is an array of the specified type (hence the generic declaration) and num_items() just returns the count of the array.

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生生漫 2024-11-17 05:14:58

CustomObject<int> co();

这是一个函数声明。省略括号。

std::cout << co();

为什么要将 operator() 应用于 co？再次省略括号。这应该可行：

CustomObject<int> co;
std::cout << co;

唉，从 print 方法构建和返回字符串并不是惯用的 C++。这是我要做的：

template <typename T>
class CustomObject
{
    // ...

public:

    void print(std::ostream& os) const
    {
        for (int i = 0; i != num_items(); ++i)
        {
            os << queue[i] << " | \n";
        }
    }
};

std::ostream& operator<<(std::ostream& os, const CustomObject& object)
{
    object.print(os);
    return os;
}

CustomObject<int> co();

That's a function declaration. Leave out the parenthesis.

std::cout << co();

Why are you appling operator() to co? Again, leave out the parenthesis. This should work:

CustomObject<int> co;
std::cout << co;

Alas, building and returning a string from a print method is hardly idiomatic C++. Here is what I would do:

template <typename T>
class CustomObject
{
    // ...

public:

    void print(std::ostream& os) const
    {
        for (int i = 0; i != num_items(); ++i)
        {
            os << queue[i] << " | \n";
        }
    }
};

std::ostream& operator<<(std::ostream& os, const CustomObject& object)
{
    object.print(os);
    return os;
}

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