C ++“未在此范围内声明变量” - 再次
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C++ "Variable not declared in this scope" - again
提问by Jubbsee
I guess this is a really simple question and, probably, one that has been answered several times over. However, I really do suck at C++ and have searched to no avail for a solution. I would really appreciate the help.
我想这是一个非常简单的问题，并且可能已经被多次回答了。但是，我真的很擅长 C++，并且搜索了无济于事的解决方案。我真的很感激你的帮助。
Basically:
基本上：
#ifndef ANIMAL_H
#define ANIMAL_H
class Animal
{
public:
void execute();
void setName(char*);
Animal();
virtual ~Animal();
private:
void eat();
virtual void sleep() = 0;
protected:
char* name;
};
class Lion: public Animal
{
public:
Lion();
private:
virtual void sleep();
};
class Pig: public Animal
{
public:
Pig();
private:
virtual void sleep();
};
class Cow: public Animal
{
public:
Cow();
private:
virtual void sleep();
};
#endif
Is the header file, where:
是头文件，其中：
#include <iostream>
#include "Animal.h"
using namespace std;
Animal::Animal()
{
name = new char[20];
}
Animal::~Animal()
{
delete [] name;
}
void setName( char* _name )
{
name = _name;
}
void Animal::eat()
{
cout << name << ": eats food" << endl;
}
void Animal::execute()
{
eat();
sleep();
}
Lion::Lion()
{
name = new char[20];
}
void Lion::sleep()
{
cout << "Lion: sleeps tonight" << endl;
}
Pig::Pig()
{
name = new char[20];
}
void Pig::sleep()
{
cout << "Pig: sleeps anytime, anywhere" << endl;
}
Cow::Cow()
{
name = new char[20];
}
void Cow::sleep()
{
cout << "Cow: sleeps when not eating" << endl;
}
is the C file. As you can see, really simple stuff, but, I get the: "error: ‘name' was not declared in this scope" whenever I try to compile.
是C文件。如您所见，非常简单的东西，但是，每当我尝试编译时，我都会收到：“错误：'name' 未在此范围内声明”。
It compiles if I comment out the setName method. Iv tried setting 'name' to public and still get the same error. I have also tried using "this->name = _name" in setName(), which results in "invalid use of ‘this' in non-member function".
如果我注释掉 setName 方法，它就会编译。我尝试将“名称”设置为公开，但仍然出现相同的错误。我还尝试在 setName() 中使用“this->name = _name”，这导致“在非成员函数中无效使用‘this’”。
I don't know what else to search for. Thanks in advance.
我不知道还能搜索什么。提前致谢。
回答by Bob Fincheimer
void setName( char* _name )
{
name = _name;
}
should be
应该
void Animal::setName( char* _name )
{
this->name = _name;
}
You need to have Animal::if you use the thisparameter. Without Animal::it thinks you are just creating a new global function called setName
Animal::如果您使用该this参数，则需要拥有。没有Animal::它认为您只是在创建一个名为的新全局函数setName
回答by Timo Geusch
The way you have written the code setNameis a free function, not a member function. For this reason the compiler can't resolve name.
您编写代码的方式setName是自由函数，而不是成员函数。因此，编译器无法解析name.
You'll have to change setName to this:
您必须将 setName 更改为：
void Animal::setName( char* _name )
{
name = _name;
}
回答by AnT
"It compiles if I comment out the setName method"
“如果我注释掉 setName 方法，它就会编译”
You don't have a "setName method" in your program (referring to the problematic definition). You defined a completely independent global function called setName, which is not a "method" of anything. If you want to define a method, i.e. a member function of a class, you have to refer to it using the class_name::method_nameformat. That would be Animal::setNamein your case.
您的程序中没有“setName 方法”（指有问题的定义）。您定义了一个名为 的完全独立的全局函数setName，它不是任何事物的“方法”。如果你想定义一个方法，即一个类的成员函数，你必须使用class_name::method_name格式来引用它。那就是Animal::setName你的情况。
回答by Douglas Leeder
The hint was "non-member function".
提示是“非成员函数”。
You need to make the function into a member function:
您需要将函数变成成员函数：
void Animal::setName( char* _name )
{
name = _name;
}
