C++ 模板中依赖类型的问题
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2931345/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Trouble with dependent types in templates
提问by Nick Heiner
I'm having trouble with templates and dependent types:
我在使用模板和依赖类型时遇到问题：
namespace Utils
{
void PrintLine(const string& line, int tabLevel = 0);
string getTabs(int tabLevel);
template<class result_t, class Predicate>
set<result_t> findAll_if(typename set<result_t>::iterator begin, set<result_t>::iterator end, Predicate pred) // warning C4346
{
set<result_t> result;
return findAll_if_rec(begin, end, pred, result);
}
}
namespace detail
{
template<class result_t, class Predicate>
set<result_t> findAll_if_rec(set<result_t>::iterator begin, set<result_t>::iterator end, Predicate pred, set<result_t> result)
{
typename set<result_t>::iterator nextResultElem = find_if(begin, end, pred);
if (nextResultElem == end)
{
return result;
}
result.add(*nextResultElem);
return findAll_if_rec(++nextResultElem, end, pred, result);
}
}
Compiler complaints, from the location noted above:
编译器投诉，来自上述位置：
warning C4346: 'std::set<result_t>::iterator' : dependent name is not a type. prefix with 'typename' to indicate a type
error C2061: syntax error : identifier 'iterator'
What am I doing wrong?
我究竟做错了什么？
回答by James McNellis
Well, the warning says:
好吧，警告说：
dependent name is not a type. prefix with 'typename' to indicate a type
依赖名称不是类型。带有“typename”的前缀以指示类型
The dependent name (that is, the iteratorin std::set<result_t>::iterator) is not a type. You need to prefix it with typenameto indicate a type:
依赖名称（即iteratorin std::set<result_t>::iterator）不是类型。您需要在它前面加上前缀typename以指示类型：
typename std::set<result_t>::iterator
So, your declaration should be:
所以，你的声明应该是：
template<class result_t, class Predicate>
set<result_t> findAll_if(typename set<result_t>::iterator begin, typename set<result_t>::iterator end, Predicate pred)
note added typename ^
(and the definition should match the declaration)
（并且定义应与声明匹配）
回答by GBegen
You need an additional typename keyword on this line:
在这一行你需要一个额外的 typename 关键字：
set<result_t> findAll_if(typename set<result_t>::iterator begin,typenameset<result_t>::iterator end, Predicate pred) // warning C4346
set<result_t> findAll_if(typename set<result_t>::iterator begin,类型名称set<result_t>::iterator end, Predicate pred) // warning C4346
