没有匹配“ operator<<”的对象在std :: cout中 [英] No match for 'operator<<' in std::cout

问题描述

我正在开发gsoap Web服务，在其中我将检索对象的向量以返回查询。我有两种方法可以做到：首先通过简单循环和迭代器。它们都不起作用。

I am developing gsoap web service where I am retrieving vectors of objects in return of a query. I have two ways to do it: first by simple loop and by iterator. None of them working.

错误是：

错误：在'std :: cout
mPer.MultiplePersons :: info.std :: vector<中没有'operator<<'的匹配项; _Tp，_Alloc> :: at< PersonInfo，std :: allocator< PersonInfo> >（（（（std :: vector< PersonInfo> :: size_type）i））'



error: no match for 'operator<<' in 'std::cout mPer.MultiplePersons::info.std::vector<_Tp, _Alloc>::at<PersonInfo, std::allocator<PersonInfo> >(((std::vector<PersonInfo>::size_type)i))'

MultiplePersons mPer; // Multiple Person is a class, containing vector<PersonInfo> info
std::vector<PersonInfo>info; // PersonInfo is class having attributes Name, sex, etc.
std::vector<PersonInfo>::iterator it;

cout << "First Name: \t";
cin >> firstname;
if (p.idenGetFirstName(firstname, &mPer) == SOAP_OK) {
    // for (int i = 0; i < mPer.info.size(); i++) {
    //    cout << mPer.info.at(i); //Error
    //}
    for (it = info.begin(); it != info.end(); ++it) {
        cout << *it; // Error
    }

} else p.soap_stream_fault(std::cerr);

}

很明显，运算符重载 operator< ; cout 中的< 是问题。我研究了与此相关的几个问题，但是没有人帮助我。如果有人可以提供有关如何解决该问题的具体示例，将不胜感激。 （请不要大声谈论它，我是C ++的新手，并且花了三天时间在它上面寻找解决方案。）

It's obvious that operator overloading operator<< in cout is the problem. I have looked at several problems related to this, but no one helped me out. If someone can provide a concrete example on how to solve it, it would be very appreciated. (Please do not talk in general about it, I am new to C++ and I have spent three days on it searching for solution.)

推荐答案

您需要为 PersonInfo 提供输出流运算符。像这样：

You need to provide an output stream operator for PersonInfo. Something like this:

struct PersonInfo
{
  int age;
  std::string name;
};

#include <iostream>
std::ostream& operator<<(std::ostream& o, const PersonInfo& p)
{
  return o << p.name << " " << p.age;
}

此运算符允许使用 A < < B ，其中 A 是 std :: ostream 实例（其中 std :: cout 是一个），而 B 是 PersonInfo 实例。

This operator allows expressions of the type A << B, where A is an std::ostream instance (of which std::cout is one) and B is a PersonInfo instance.

这允许您执行以下操作：

This allows you do do something like this:

#include <iostream>
#include <fstream>
int main()
{
  PersonInfo p = ....;
  std::cout << p << std::endl; // prints name and age to stdout

  // std::ofstream is also an std::ostream, 
  // so we can write PersonInfos to a file
  std::ofstream person_file("persons.txt");
  person_file << p << std::endl;
}

反过来，您可以打印取消引用的迭代器。

which in turn allows you to print the de-referenced iterator.

这篇关于没有匹配“ operator<<”的对象在std :: cout中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！