C++类 - 如何从另一个成员函数引用成员函数

发布于 2024-10-30 23:38:28 字数 452 浏览 1 评论 0原文

我对 C++ 类非常陌生，所以这可能是一个非常明显的问题，但因为我不熟悉行话，但我似乎无法获得正确的搜索术语。

不管怎样，我想做的是让类中的公共函数访问同一个类中的私有函数。

例如

//.h file:

class foo {

float useful(float, float);

public:

int bar(float);

};

//.cpp file:

int foo::useful(float a, float b){
//does something and returns an int
}

int foo::bar(float a){
//how do I access the 'useful' function in foo?? eg something like
return useful(a, 0.8); //but this doesnt compile
}

原文

I'm very new to c++ classes, so this is probably a really obvious question, but because I'm not familiar with the lingo yet I cant seem to be able to get a correct search term.

Anyway, what I am trying to do is have a public function in a class access a private function in same class.

//.h file:

class foo {

float useful(float, float);

public:

int bar(float);

};

//.cpp file:

int foo::useful(float a, float b){
//does something and returns an int
}

int foo::bar(float a){
//how do I access the 'useful' function in foo?? eg something like
return useful(a, 0.8); //but this doesnt compile
}

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寻梦旅人 2024-11-06 23:38:28

函数 useful 被声明为返回 float，但您将其定义为返回 int。

对比

float useful(float, float);

与

int foo::useful(float a, float b){
    //does something and returns an int
}

如果将声明更改为 int important(float, float) 并从函数中返回一些内容，它将正常工作。

The function useful is declared to return a float, but you define it as returning an int.

Contrast

float useful(float, float);

int foo::useful(float a, float b){
    //does something and returns an int
}

If you change the declaration to int useful(float, float) and return something from the function it will work fine.

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不打扰别人 2024-11-06 23:38:28

您的返回类型不匹配：

//.h file:

class foo {

float useful(float, float);      // <--- THIS ONE IS FLOAT ....

public:

int bar(float);

};

//.cpp file:

int foo::useful(float a, float b){       // <-- ...THIS ONE IS INT. WHICH ONE?
//does something and returns an int
}

int foo::bar(float a){
//how do I access the 'useful' function in foo?? eg something like
return useful(a, 0.8); //but this doesnt compile
}

编译器会查找完全匹配的函数定义。您收到的编译器错误可能是在抱怨以下事实：a) 找不到 floatUsed()，或者 b) 当您谈论 < 时不知道您的意思代码>int有用。

确保这些匹配，并且在 bar 中调用 useful 应该可以正常工作。

Your return types don't match:

//.h file:

class foo {

float useful(float, float);      // <--- THIS ONE IS FLOAT ....

public:

int bar(float);

};

//.cpp file:

int foo::useful(float a, float b){       // <-- ...THIS ONE IS INT. WHICH ONE?
//does something and returns an int
}

int foo::bar(float a){
//how do I access the 'useful' function in foo?? eg something like
return useful(a, 0.8); //but this doesnt compile
}

The compiler looks for a function definition that matches exactly. The compiler error that you are getting is probably complaining about the fact that it a) can't find float useful(), or b) doesn't know what you mean when you're talking about int useful.

Make sure those match, and calling useful within bar should work just fine.

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