升压Gzip已过滤器：编译failes

Question

我试图编译例如从升压Gzip已过滤器页面：

#include <fstream> 
#include <iostream> 
#include <boost/iostreams/filtering_streambuf.hpp> 
#include <boost/iostreams/copy.hpp> 
#include <boost/iostreams/filter/gzip.hpp> 

int main() 
{ 
    using namespace std; 

    ifstream file("hello.gz", ios_base::in | ios_base::binary); 
    filtering_streambuf<input> in; 
    in.push(gzip_decompressor()); 
    in.push(file); 
    boost::iostreams::copy(in, cout); 
} 

可悲的是我的G ++会返回错误：

gzlib.cpp: In function ‘int main()’: 
gzlib.cpp:12:3: error: ‘filtering_streambuf’ was not declared in this scope 
gzlib.cpp:12:23: error: ‘input’ was not declared in this scope 
gzlib.cpp:12:30: error: ‘in’ was not declared in this scope 
gzlib.cpp:13:29: error: ‘gzip_decompressor’ was not declared in this scope 

有什么不对的功能，以及如何修改它使其工作？非常感谢！

链接，以提高Gzip已过滤器：http://www.boost.org/doc/libs/release/libs/iostreams/doc/classes/gzip.html

Answer 1

的问题是，你没有指定要在其中查找filtering_streambuf，input，或gzip_decompressor命名空间。 尝试：

#include <fstream> 
#include <iostream> 
#include <boost/iostreams/filtering_streambuf.hpp> 
#include <boost/iostreams/copy.hpp> 
#include <boost/iostreams/filter/gzip.hpp> 

int main() 
{ 
    using namespace std; 
    using namespace boost::iostreams; 
    ifstream file("hello.gz", ios_base::in | ios_base::binary); 
    filtering_streambuf<input> in; 
    in.push(gzip_decompressor()); 
    in.push(file); 
    copy(in, cout); 
} 

---

的原因，example不这样做，这是因为建立在introduction公约：

文档中介绍的所有类，功能和模板是命名空间boost :: iostreams，除非另有说明。命名空间限定通常被忽略。