返回指针的模板函数出错 [英] Error on template functions returning pointers

问题描述

我正在尝试模拟基于指针的列表类和几个函数（我尝试返回指针的函数）正在生成错误消息。


班级：

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解决方案

< blockquote>尝试在ListNode之后向右移动*。所以，ListNode * List< T> :: getHead（），而不是ListNode * List< T> :: getHead（）。第二个意味着该类被称为* List< T>并且该函数返回ListNode。第一个告诉编译器该类被称为List< T>并且该函数返回一个ListNode指针。

嗯......感谢您的建议，但它似乎没有什么区别。还有其他想法吗？


再次感谢！

尝试在ListNode之后向右移动* 。所以，ListNode * List< T> :: getHead（），而不是ListNode * List< T> :: getHead（）。第二个意味着该类被称为* List< T>并且该函数返回ListNode。第一个告诉编译器该类被称为List< T>并且该函数返回ListNode指针。

Laharl，实际上，这里的空白是一个偏好问题 - 特别是因为类名不能以*字符开头（否则每次你说int * x时，你会得到一个错误）。


我希望，由于ListNode本身就是一个模板结构，你需要将它作为返回类型的模板：

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I''m trying to template a pointer-based list class and several functions (the ones in which I try to return a pointer) are producing error messages.

The class:

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解决方案

Try moving the * to right after ListNode. So, ListNode* List<T>::getHead(), rather than ListNode *List<T>::getHead(). The second one implies that the class is called *List<T> and the function returns a ListNode. The first tells the compiler that the class is called List<T> and the function returns a ListNode pointer.

Hmm... Thanks for the suggestion, but it didn''t seem to make a difference. Any other ideas?

Thanks again!

Try moving the * to right after ListNode. So, ListNode* List<T>::getHead(), rather than ListNode *List<T>::getHead(). The second one implies that the class is called *List<T> and the function returns a ListNode. The first tells the compiler that the class is called List<T> and the function returns a ListNode pointer.

Laharl, actually, the whitespace here is a matter of preference - especially since classnames cannot start with the * character (else every time you said int *x, you would get an error).

I expect that, since ListNode is itself a template struct, you need to treat it as a template in your return type:

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