问题描述
我在一个类中使用了几个函数,这些函数通过函数接口传递了一个 ostream,然后可以用来输出错误消息。我曾希望能够将所有 ostream 绑定到一个对象,然后在必要时重定向到一个文件。
我的代码的相关部分如下所示:
#include <iostream>
class Example
{
public:
Example(){} //<--Error: "std::basic_ostream<_CharT,_Traits>::basic_ostream() [with _CharT=char,_Traits=std::char_traits<char>]" (declared at line 390 of "/usr/include/c++/9/ostream") is inaccessible C/C++(330)
void DoSomething()
{
FunctionWithOstream(out);
}
private:
std::ostream out; //in my case,the ostream is currently not needed for the time being.
void FunctionWithOstream(std::ostream& out)
{
out << "Something";
}
};
在构造函数(或程序中的所有构造函数)的第一个大括号处,我收到以下错误消息:
受保护的函数“std::basic_ostream<_chart _traits>::basic_ostream() [with _CharT=char,_Traits=std::char_traits]”(在“/usr/include/c++/9/ostream”的第 390 行声明)不能通过
"std::basic_ostream
"std::basic_ostream<_chart _traits>::basic_ostream() [与 _CharT=char,_Traits=std::char_traits]”(在“/usr/include/c++/9/ostream”的第 390 行声明)不可访问 C/C++(330)
我希望问题足够清楚,并提前感谢您抽出宝贵时间。
您好 提尔曼
解决方法
std::ostream
不是默认可构造的,您可能需要引用/指针:
class Example
{
public:
Example(std::ostream& out = std::cout) : out(out) {}
void DoSomething() { FunctionWithOstream(out); }
private:
std::ostream& out;
void FunctionWithOstream(std::ostream& os) { os << "Something"; }
};
,
好的,我找到了一个解决方案:
#include <iostream>
#include <sstream>
class Example
{
public:
Example(){} //<--Error: "std::basic_ostream<_CharT,_Traits>::basic_ostream() [with _CharT=char,_Traits=std::char_traits<char>]" (declared at line 390 of "/usr/include/c++/9/ostream") is inaccessible C/C++(330)
void DoSomething()
{
FunctionWithOstream(outStream);
}
private:
std::ostringstream outStream; //std::ostringstream instead of std::ostream is working fine.
void FunctionWithOstream(std::ostream& out)
{
out << "Something";
}
};
可以使用 std::ostream 代替 std::ostream 来接收函数流。