C ++模板类错误与operator == [英] C++ template class error with operator ==

问题描述

错误：

错误C2678：binary'=='：无操作符，它接受类型为const entry的左侧操作数（或没有可接受的转换）

Error:
error C2678: binary '==' : no operator found which takes a left-hand operand of type 'const entry' (or there is no acceptable conversion)

函数：

template <class T, int maxSize>
int indexList<T, maxSize>::search(const T& target) const
{
    for (int i = 0; i < maxSize; i++)  
    	if (elements[i] == target)   //ERROR???
    		return i;       // target found at position i

    // target not found
    return -1;
}

indexList.h

indexList.cpp

这假设是一个重载运算符？作为模板类我不知道我是否理解错误？

Is this suppose to be an overloaded operator? Being a template class I am not sure I understand the error?

解决方案 -
类中的重载函数现在声明了const：

Solution- The overload function in the class now declared const:

//Operators
bool entry::operator == (const entry& dE)  const <--
{
    return (name ==dE.name);

}

推荐答案

正确地读取错误文本：

二进制'=='：无操作符找到一个左手操作数类型' const entry'

binary '==' : no operator found which takes a left-hand operand of type 'const entry'

这意味着它找不到任何 == 接受条目类型作为其左操作数。此代码无效：

It means it can't find any == operator that accepts an entry type as its left operand. This code isn't valid:

entry const e;
if (e == foo)

类，但这不是错误是什么。错误是关于条目类型缺乏运算符，无论是什么。或者给类一个运算符== 函数，或者声明一个独立的运算符== 函数，接受 const entry& 作为其第一个参数。

You've showed us the code for your list class, but that's not what the error is about. The error is about a lack of operators for the entry type, whatever that is. Either give the class an operator== function, or declare a standalone operator== function that accepts a const entry& as its first parameter.

struct entry {
  bool operator==(const entry& other) const;
};
// or
bool operator==(const entry& lhs, const entry& rhs);

我认为后者是首选风格。

I think the latter is the preferred style.

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