C++将对象添加到矢量时出错
我是矢量新手。我正在尝试将对象添加到向量中。但是程序无法编译，因为我的代码有问题。但我不知道那是什么。错误是：
error C2664: 'void std::vector<_Ty>::push_back(_Ty &&)' : cannot convert parameter 1 from 'Line (void)' to 'Line &&'
代码是：
Line help_line ();
cin >> ln_quan;
vector <Line> figure_line;
for (int i = 0 ; i < ln_quan ; i++)
{
figure_line.push_back(help_line);
}
编译器说错误在第 6 行（figure_line.push_back(help_line);）。
我放弃了试图找到一个解释如何添加对象的教程（在做这样的事情时我很容易放弃......）。
'Line (void)' 和 'Line &&' 是什么意思意思是？ “Line (void)”是“Line”类吗？如果是这样，在这种情况下“(void)”是什么意思？
I'm new with vectors. I'm trying to add objects to a vector. But the program can't compile because I have a problem in the code. But I don't know what is it. The error is:
error C2664: 'void std::vector<_Ty>::push_back(_Ty &&)' : cannot convert parameter 1 from 'Line (void)' to 'Line &&'
The code is:
Line help_line ();
cin >> ln_quan;
vector <Line> figure_line;
for (int i = 0 ; i < ln_quan ; i++)
{
figure_line.push_back(help_line);
}
The compiler says that the error is at the 6-th line (figure_line.push_back(help_line);).
I gave up trying to find a tutorial explaining how to add objects (I give up easily when doing such things...).
And what does 'Line (void)' and 'Line &&' mean? Is 'Line (void)' the class 'Line'? If so, what does '(void)' mean in this case?
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这声明了一个函数，而不是一个
Line。请改用Line help_line;。请参阅：最令人烦恼的解析：为什么不 A a(());工作？
This declares a function, not a
Line. UseLine help_line;instead.See: Most vexing parse: why doesn't A a(()); work?
您已将
help_line声明为不带参数并返回Line的函数。这就是你的意图吗？如果是这样，那么您需要调用该函数，如下所示：
如果不是，并且您打算将
help_line声明为Line类型的对象，则需要这样：You have declared
help_lineas a function taking no parameters and returning aLine. Is that what you intended?If so, then you need to invoke the function, like this:
If not, and you intended to declare
help_lineas an object of typeLine, you need this:这并不意味着“
help_line应该是使用默认构造函数创建的Line实例”。这意味着“help_line应该是一个函数，在其他地方实现，不带参数并返回一个Line实例”。您想要的内容拼写为
Line help_line;，不带括号。因此，您会收到以下错误消息：
Line &&是push_back所期望的参数类型。&&在这里并不重要；对于初学者来说，最好将其视为一种调用约定。您仍然只是传递一个Line，因为这是您在Line向量中收集的内容。Line(void)是“不带参数并返回Line实例的函数类型”。(void)是函数参数的另一种编写()的方法（在新代码中不鼓励这样做，但有时在与非常旧的 C 代码交互时需要）。This does not mean "
help_lineshall be an instance ofLinecreated with the default constructor". It means "help_lineshall be a function, implemented somewhere else, that takes no arguments and returns aLineinstance".The thing you want is spelled
Line help_line;, with no parentheses.So, you get the following error message:
Line &&is the kind of parameter thatpush_backis expecting. The&&doesn't really matter here; it's best thought of, for beginners, as a kind of calling convention. You're still just passing aLine, because that's the kind of thing you collect in a vector ofLines.Line(void)is "the type of functions that take no arguments and return aLineinstance".(void)is another way to write(), for function arguments (it is discouraged in new code, but sometimes needed when interacting with very old C code).