C++ '.' 之前的预期主要表达式

Expected primary expression before '.'

提问by photon

#include <iostream>
using std::cout;
using std::endl;

class square {

public:
    double length, width;

    square(double length, double width);
    square();

    ~square();

    double perimeter();
};

double square::perimeter() {
return 2*square.length + 2*square.width;
}

int main() {

square sq(4.0, 4.0);

cout << sq.perimeter() << endl;

return 0;
}

I'm trying to write a simple class program. I am getting the error

我正在尝试编写一个简单的类程序。我收到错误

in member function 'double square::perimeter()'; .cpp:21: error: expected primary-expression before '.' token .cpp:21: error: expected primary-expression before '.' token

在成员函数'double square::perimeter()'中；.cpp:21: 错误：'.' 之前的预期主表达式 标记 .cpp:21: 错误：'.' 之前的预期主表达式 令牌

Does this mean I have to use 2*square::length + 2*square::width??

这是否意味着我必须使用 2*square::length + 2*square::width？

回答by ildjarn

squareis a type, not an object; instead of

square是一个类型，而不是一个对象；代替

return 2*square.length + 2*square.width;

do

做

return 2*length + 2*width;

(which is implicitly the same as:

（这与以下内容隐含相同：

return 2*this->length + 2*this->width;

which you may, or ^pleasemay not, prefer for clarity).

你可以，或者^请可能不会，更喜欢清晰）。

2*square::length + 2*square::widthwould be valid syntax if lengthand widthwere

2*square::length + 2*square::width如果length并且width是，将是有效的语法

• staticmembers of square, or
• members of some base class square, or
• objects in some namespace square

• 的静态成员square，或
• 某些基类的成员square，或
• 某个命名空间中的对象 square

回答by Lightness Races in Orbit

Yes, the accurate form would be:

是的，准确的形式是：

return 2*square::length + 2*square::width;

since squareis a type, not an object.

因为square是类型，而不是对象。

In this context, it's the same as:

在这种情况下，它与以下内容相同：

return 2*this->square::length + 2*this->square::width;

However, since it's the sameobject and the sametype, you can just write:

但是，由于它是相同的对象和相同的类型，你可以只写：

return 2*this->length + 2*this->width;

or, most simply:

或者，最简单的是：

return 2*length + 2*width;

回答by Gordon Gustafson

double square::perimeter() {
    return 2*square.length + 2*square.width;
}

You need to say square::perimeter()because you are defining a method of the square class itself. It may seem like you want to define it on a specific object, but you want it to be available to all instances of square, so you need to define it on a specific one.

你需要说，square::perimeter()因为你正在定义square类本身的一个方法。看起来您想在特定对象上定义它，但您希望它可用于所有 square 实例，因此您需要在特定对象上定义它。

The instance variables lengthand widthon the other hand, pertain to a specific instance of a class and NOT the entire class itself (otherwise you would declare them as static). This means that you can just refer to them directly instead of going through the class itself, and the compiler will know what variables you're talking about. This is because widthand lengthare defined in the same scope as the method, so you don't need to give it special directions with ::to tell it where to find what its looking for. Hence:

实例变量length，并width在另一方面，属于一类的特定实例而不是整个类本身（否则你将宣布他们为static）。这意味着您可以直接引用它们而不是通过类本身，并且编译器将知道您在谈论哪些变量。这是因为width和length定义在与方法相同的范围内，所以您不需要给它特殊的指示::来告诉它在哪里可以找到它正在寻找的东西。因此：

double square::perimeter() {
    return 2 * length + 2 * width; //will refer to the instance variables automatically
}

回答by Bart

Simply use

只需使用

double square::perimeter() {
    return 2 * length + 2 * width;
}

回答by Karoly Horvath

to access instance variables, just name them:

要访问实例变量，只需命名它们：

double square::perimeter() {
  return 2*length + 2*width;
}

回答by Rohit Vipin Mathews

double square::perimeter() {
return 2*square.length + 2*square.width;
}

what is the squarein this function? square is a class. you use the .operator to acces members from objects. like sq.somefun();

什么是square这个功能呢？square 是一个类。您可以使用.运算符访问对象中的成员。喜欢sq.somefun();

so you perimeter function should be:

所以你的周长函数应该是：

 double square::perimeter() {
    return (2*(length + width));
    }

But isnt the length and width of square equal?

但是正方形的长和宽不一样吗？