如何在C ++中返回NULL对象 [英] How to return NULL object in C++

问题描述

我知道这可能是一个重复的： C ++返回NULL对象，如果没有找到搜索结果

I know that this might be a duplicate of: C++ return a "NULL" object if search result not found

但是我的代码有一些不同的地方，因为星号不能解决我的问题，这是： / p>

BUT, there's something different going on with my code because the asterisk doesn't solve my problem, which is this:

Normal Sphere::hit(Ray ray) {
   //stuff is done here
   if(something happens) {
       return NULL;
   }
   //other stuff
   return Normal(something, somethingElse);
}

但是我得到一个错误引用 / code> line：从'int'转换为非标量类型'正常'请求

But I get an error referencing the return NULL line: conversion from ‘int’ to non-scalar type ‘Normal’ requested

另一个错误和警告引用了最后一个返回行：警告：取地址临时和从正常*转换为非标量类型'正常请求

And another error and warning that referencing the last return line: warning: taking address of temporary and conversion from ‘Normal*’ to non-scalar type 'Normal' requested

我理解为什么我收到此警告，但我不知道如何解决它。如何在函数结束后在最后一行返回一个 Normal 对象，如何返回 NULL 对象第一次？ （如果有这些类型的退货的条款，请让我知道，所以我也可以读更多。）

I understand why I am getting this warning, but I don't know how to fix it. How do I return a Normal object in the last line that persists after the function ends and how do I return a NULL object that first time? (If there's a term for these types of returns, please let me know so I can also read up on it more.)

为了澄清评论者的问题，我尝试这些东西：

To clarify a commenter's question, I've tried these things:

我试着这样做： Normal * Sphere :: hit（Ray ray）文件和正常*命中（Ray ray）; 在头文件中，我得到这个错误： （Ray）'在类'Sphere'中不匹配

I tried doing this: Normal *Sphere::hit(Ray ray) in the cpp file and Normal *hit( Ray ray ); in the header file and I get this error: error: prototype for ‘Normal* Sphere::hit(Ray)’ does not match any in class 'Sphere'

我也尝试过： *在头文件中的cpp文件和正常*命中（Ray ray）; 命中（Ray ray）第二个return语句：无法将返回中的Normal *转换为Normal Sphere :: *

I also tried this: Normal Sphere::*hit(Ray ray) in the cpp file and Normal *hit( Ray ray); in the header file and I get this error for the second return statement: cannot convert 'Normal*' to 'Normal Sphere::*' in return

进一步说明：我不是问指针是如何工作的。 （这不是主要问题。）我想知道关于C ++中的指针的语法。所以，给定我上面指定的函数，我收集了我应该指定一个返回一个指针，因为C ++没有空对象。得到它了。但是，问题就变成了：函数原型应该是什么样子？在cpp文件中，我有Bala建议（这是我原来，但更改，因为以下错误）：

Further clarification: I'm not asking about how pointers work. (That wasn't the main question.) I'm wondering about syntax regarding pointers in C++. So, given the function I've specified above, I've gleaned that I should specify a return a pointer because C++ doesn't have null objects. Got it. BUT, the problem then becomes: what should the function prototype look like? In the cpp file, I have what Bala suggested (which is what I had originally but changed it because of the following error):

Normal* Sphere::hit(Ray ray) {
   //stuff is done here
   if(something happens) {
       return NULL;
   }
   //other stuff
   return new Normal(something, somethingElse);
}

在头文件中，我有 Normal * hit （Ray ray），但我仍然得到这个消息：'Normal * Sphere :: hit（Ray）'的原型与类'Sphere' c $ c>在这一点上，我不清楚为什么它找不到该函数原型。这是头文件：

In the header file, I have Normal *hit(Ray ray), but I still get this message: prototype for 'Normal* Sphere::hit(Ray)' does not match any in class 'Sphere' At this point, it is unclear to me why it can't find that function prototype. Here is the header file:

class Sphere
{
    public:
        Sphere();
        Vector3 center;
        float radius;
        Normal* hit(Ray ray);
};

任何人都能看到为什么抱怨没有匹配的原型在 Sphere 类中命中？ （我可能会将此移动到一个单独的问题...）

Can anyone see why it's complaining that there doesn't exist a matching prototype for hit in the Sphere class? (I might move this to a separate question...)

推荐答案

我认为你需要像

Normal* Sphere::hit(Ray ray) {
   //stuff is done here
   if(something happens) {
       return NULL;
   }
   //other stuff
   return new Normal(something, somethingElse);
}

才能返回NULL;

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