C++“...不命名类型”

C++ "...does not name a type"

我一直在尝试定义一个 class 方法,它使用在 class 命名空间中声明的 return 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,出现此错误:

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

我也试过这样定义方法:

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

我做错了什么?

名称 iterator 的范围仅限于您的 class,并且它是从属名称。为了使用它,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 M.M 在评论中指出的,您还可以使用尾随 return 语法作为

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

Live Example