C++ C++；与操作员不匹配<&书信电报；

这是我的代码，我不知道为什么会抛出错误-该方法应该对指针对象进行操作并打印出其值

主要内容：

---

cout您要使用的错误很明显

cout << "Deleted item is: " << Priority_Queue::displayRecord(tmp);

在函数调用之前添加
Priority\u Queue:：
有什么原因吗？嗯，我认为是命名空间，OP没有在发布的代码中显示。Priority\u Queue：：是不必要的。
#include <iostream>
using namespace std;
struct PRecord {
long time;
string entry;
struct PRecord *link;
};

void displayRecord(PRecord* pr);

 error: no match for 'operator<<' in 'std::operator<< [with _Traits = std::char_traits<char>](((std::basic_ostream<char, std::char_traits<char> >&)(& std::cout)), ((const char*)"Deleted item is: ")) << Priority_Queue::displayRecord(tmp)'
    /usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/ostream:108: note: candidates are: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>& (*)(std::basic_ostream<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]
    /usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/ostream:117: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ios<_CharT, _Traits>& (*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT = char, _Traits = std::char_traits<char>]
    /usr/lib/gcc/x86_64-redhat-linux/4.4.7/../../../../include/c++/4.4.7/ostream:127: note:                 std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>]
    /usr/lib/gcc/x86_64-redhat-linux

cout << "Deleted item is: " << Priority_Queue::displayRecord(tmp);

cout << "Deleted item is: ";
Priority_Queue::displayRecord(tmp);