c++ - 操纵器是否以某种方式转换流类型？

Question

我正在尝试使用匿名ostringstream来生成string：Use an Anonymous Stringstream to Construct a String

但是，当我使用操纵器时，我似乎无法再编译：

const auto myString(static_cast<ostringstream>(ostringstream{} << setfill('!') << setw(13) << "lorem ipsum").str());

但即使在 gcc 5.1 中似乎也不允许这样做：

prog.cpp: In function int main():
prog.cpp:8:109: error: no matching function for call to std::basic_ostringstream<char>::basic_ostringstream(std::basic_ostream<char>&)
const auto myString(static_cast<ostringstream>(ostringstream{} << setfill('!') << setw(13) << "lorem ipsum").str());


In file included from /usr/include/c++/5/iomanip:45:0,
from prog.cpp:1:
/ usr/include/c++/5/sstream:582:7: 注意：候选
std::basic_ostringstream<_CharT, _Traits, _Alloc>::basic_ostringstream(std::basic_ostringstream<_CharT, _Traits, _Alloc>&&)[with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>]
basic_ostringstream(basic_ostringstream&& __rhs)

/usr/include/c++/5/sstream:582:7: 注意：没有已知的参数 1 从std::basic_ostream<char>到std::basic_ostringstream<char>&&
/usr/include/c++/5/sstream:565:7 的转换：注意：候选：
std::basic_ostringstream<_CharT, _Traits, _Alloc>::basic_ostringstream(const __string_type&, std::ios_base::openmode)[with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>; std::basic_ostringstream<_CharT, _Traits, _Alloc>::__string_type = std::basic_string<char>; std::ios_base::openmode = std::_Ios_Openmode]
basic_ostringstream(const __string_type& __str,

/usr/include/c++/5/sstream:565:7: 注意：没有已知的参数 1 从std::basic_ostream<char>到const __string_type& {aka const std::basic_string<char>&}
/usr/include/c++/5/sstream:547:7 的转换：注意：候选：
std::basic_ostringstream<_CharT, _Traits, _Alloc>::basic_ostringstream(std::ios_base::openmode)[with_CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>; std::ios_base::openmode = std::_Ios_Openmode]
basic_ostringstream(ios_base::openmode __mode = ios_base::out)

/usr/include/c++/5/sstream:547:7: 注意：没有已知的参数 1 从std::basic_ostream<char>到的转换std::ios_base::openmode {aka std::_Ios_Openmode}

这是另一个 gcc 流错误，还是我所做的实际上是非法的？

Answer 1

static_cast<ostringstream>(...)

这将尝试ostringstream从括号中的参数a 构造一个 new std::ostream&，其中没有 的构造函数std::ostringstream。

您只想将引用转换回原始类型。将演员转换为参考：

static_cast<ostringstream&>(...)

然后它工作正常。

我不知道您认为什么有效，但是省略了参考并删除了操纵器，它仍然失败。