C++"......没有命名类型"

Question

我一直在尝试定义一个类方法,它使用在类命名空间中声明的返回类型:

template<class T, int SIZE>
class SomeList{

public:

    class SomeListIterator{
        //...
    };

    using iterator = SomeListIterator;

    iterator begin() const;

};

template<class T, int SIZE>
iterator SomeList<T,SIZE>::begin() const {
    //...
}

当我尝试编译代码时,我收到此错误:

Building file: ../SomeList.cpp
Invoking: GCC C++ Compiler
g++ -std=c++0x -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"SomeList.d" -MT"SomeList.d" -o "SomeList.o" "../SomeList.cpp"
../SomeList.cpp:17:1: error: ‘iterator’ does not name a type
 iterator SomeList<T,SIZE>::begin() const {
 ^
make: *** [SomeList.o] Error 1

我也尝试定义这样的方法:

template<class T, int SIZE>
SomeList::iterator SomeList<T,SIZE>::begin() const {
    //...
}

还有这个:

template<class T, int SIZE>
SomeList<T,SIZE>::iterator SomeList<T,SIZE>::begin() const {
    //...
}

结果:

Building file: ../SomeList.cpp
Invoking: GCC C++ Compiler
g++ -std=c++0x -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"SomeList.d" -MT"SomeList.d" -o "SomeList.o" "../SomeList.cpp"
../SomeList.cpp:17:1: error: invalid use of template-name ‘SomeList’ without an argument list
 SomeList::iterator SomeList<T,SIZE>::begin() const {
 ^
make: *** [SomeList.o] Error 1

我究竟做错了什么？

Answer 1

该名称iterator的范围是您的类,它是一个从属名称.要使用它,您需要使用范围运算符和typename关键字

typename SomeList<T,SIZE>::iterator SomeList<T,SIZE>::begin() const

Live Example

正如MM的评论中所指出的,你也可以使用尾随返回语法

auto SomeList<T,SIZE>::begin() const -> iterator {

Live Example