c++ - "This declaration has no storage class or type specifier in C++"

Question

我对自己的类和运算符 = 有疑问。当我尝试将 Cow 类的一个对象归因于某个函数之外的同一类的另一个对象时，我得到一条信息“此声明在 C++ 中没有存储类或类型说明符”。有什么问题？我为我的英语感到抱歉。

    #include "Header.h"

Cow cow5;
Cow cow6;
cow5 = cow6;

int main()
{

    Cow cow1;
    Cow cow2("cowa22", "hobby", 8);
    Cow cow3 = cow2;
    Cow cow4;
    cow2.operator=(cow3);
}

---

#include "Header.h"
#include <string>
#include <iostream>

Cow::Cow()
{
    strcpy_s(name, sizeof(char)*20, "unnamed");
    hobby = nullptr;
    weight = 0;
}

Cow::Cow(const char * nm, const char * ho, double wt)
{
    strcpy_s(name, sizeof(char) * 20, nm);
    hobby = new char[strlen(ho) + 1];
    strcpy_s(hobby, sizeof(char)*(strlen(ho)+1), ho);
    weight = wt;
}

Cow::Cow(const Cow & c)
{
    strcpy_s(name, sizeof(char) * 20, c.name);
    hobby = new char[strlen(c.hobby) + 1];
    strcpy_s(hobby, sizeof(char)*(strlen(c.hobby) + 1), c.hobby);
    weight = c.weight;
}

Cow::~Cow()
{
    delete[] hobby;
}

Cow & Cow::operator=(const Cow & c)
{
    if (&c == this)
        return *this;
    delete[] hobby;
    strcpy_s(name, sizeof(char) * 20, c.name);
    hobby = new char[strlen(c.hobby) + 1];
    strcpy_s(hobby, sizeof(char)*(strlen(c.hobby) + 1), c.hobby);
    weight = c.weight;

    return *this;
}

void Cow::ShowCow() const
{
    std::cout << "Name: " << name << std::endl
        << "Hobby: " << hobby << std::endl 
        << "Weight: " << weight << std::endl;

}

Answer 1

在 C++ 中，代码只能出现在函数体内或变量初始化器中。这:

cow5 = cow6;

都不在里面，所以这是一个错误。您不能让代码“ float ”在函数之外。将它放在 main 中。