为什么std :: endl生成这个神秘的错误消息? [英] Why is std::endl generating this cryptic error message?
问题描述
如果我尝试编译以下代码,我得到以下编译器错误(见代码)。如果 std :: endl
被删除,它编译没有错误。 >
#include< iostream>
#include< sstream>
#include< utility>
命名空间详细信息
{
template< class T>
void print(std :: ostream& stream,const T& item)
{
stream<项目;
}
template< class Head,class ... Tail>
void print(std :: ostream& stream,const Head& head,Tail& ... tail)
{
detail :: print(stream,head);
detail :: print(stream,std :: forward< Tail>(tail)...);
}
}
template< class ... Args>
void print(std :: ostream& stream,Args& ... args)
//注意:候选函数不可行:需要3个参数,但提供了4个
{
std :: stringstream ss;
detail :: print(ss,std :: forward< Args>(args)...);
stream<< ss.rdbuf();
}
int main()
{
print(std :: cout,答案是,42,std :: endl);
//错误:没有匹配的函数调用'print'
}
std :: endl
是一个函数模板。使用时,其模板参数必须由编译器显式指定或推导。
std :: ostream
有一个重载:
basic_ostream< charT,traits> operator< <(
basic_ostream< charT,traits>&(* pf)(basic_ostream< charT,traits>&
当我们使用
std :: cout<< std :: endl;
编译器推导出用于 std :: endl
。因为你在调用
print
时没有能力自动减少类型,所以你必须明确指出 std ::
> print(std :: cout,The answer is,42,std :: endl< char,std :: char_traits< char>>
更新
我使用以下剥离代码来跟踪问题:
#include< iostream>
命名空间详细信息
{
template< class T>
void print(std :: ostream& stream,const T& item)
{
stream<项目;
}
}
int main()
{
// detail :: print(std :: cout,std :: endl);
detail :: print(std :: cout,std :: endl< char,std :: char_traits< char>>);
}
If I try to compile the following code I get the following compiler error (see code.) It compiles without error if std::endl
is removed.
#include <iostream>
#include <sstream>
#include <utility>
namespace detail
{
template <class T>
void print(std::ostream& stream, const T& item)
{
stream << item;
}
template <class Head, class... Tail>
void print(std::ostream& stream, const Head& head, Tail&&... tail)
{
detail::print(stream, head);
detail::print(stream, std::forward<Tail>(tail)...);
}
}
template <class... Args>
void print(std::ostream& stream, Args&&... args)
//note: candidate function not viable: requires 3 arguments, but 4 were provided
{
std::stringstream ss;
detail::print(ss, std::forward<Args>(args)...);
stream << ss.rdbuf();
}
int main()
{
print(std::cout, "The answer is ", 42, std::endl);
//error: no matching function for call to 'print'
}
std::endl
is a function template. When it is used, its template parameters have to be explicitly specified or deduced by the compiler.
std::ostream
has an overload:
basic_ostream<charT,traits>& operator<<(
basic_ostream<charT,traits>& (*pf) (basic_ostream<charT,traits>&) );
When we use
std::cout << std::endl;
the compiler deduces the types to be used for std::endl
. Since you don't have the ability to fall back on automatic type deduction when calling print
, you have to be explicit about which version of std::endl
you want to use.
The following should work:
print(std::cout, "The answer is ", 42, std::endl<char, std::char_traits<char>>);
Update
I used the following stripped down code to track the issue:
#include <iostream>
namespace detail
{
template <class T>
void print(std::ostream& stream, const T& item)
{
stream << item;
}
}
int main()
{
// detail::print(std::cout, std::endl);
detail::print(std::cout, std::endl<char, std::char_traits<char>>);
}
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