将 boost 的缓冲流转换为 istream
Convert boost`s bufferstream to istream
是否可以将 boost 的缓冲流转换为 istream?我正在尝试进行转换,但是我仍然不清楚我是做错了什么还是根本不可能这样做。我将不胜感激任何答案。
char *copy = static_cast <char*> (region.get_address());
for (int i = 0;i < length;i++) copy[i] = str[i];
bufferstream input_stream (copy, length);
然后我需要将 bufferstream 转换成 istream。
基本上,我需要将 bufferstream 实例作为参数传递给接受 istream & 的函数。
不清楚你想达到什么¹,这是我最好的猜测:
#include <boost/interprocess/mapped_region.hpp>
#include <boost/interprocess/shared_memory_object.hpp>
#include <boost/interprocess/streams/bufferstream.hpp>
#include <iostream>
namespace bip = boost::interprocess;
int main() {
bip::shared_memory_object smo(bip::open_or_create, "MySharedMemory", bip::read_write);
std::string str = "test data";
smo.truncate(10ull << 10); // 10 KiB
bip::mapped_region r(smo, bip::read_write);
bip::bufferstream stream(reinterpret_cast<char*>(r.get_address()), r.get_size());
if (stream << str)
std::cout << "Written";
}
¹ https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem
² Coliru 不支持共享内存
是否可以将 boost 的缓冲流转换为 istream?我正在尝试进行转换,但是我仍然不清楚我是做错了什么还是根本不可能这样做。我将不胜感激任何答案。
char *copy = static_cast <char*> (region.get_address());
for (int i = 0;i < length;i++) copy[i] = str[i];
bufferstream input_stream (copy, length);
然后我需要将 bufferstream 转换成 istream。 基本上,我需要将 bufferstream 实例作为参数传递给接受 istream & 的函数。
不清楚你想达到什么¹,这是我最好的猜测:
#include <boost/interprocess/mapped_region.hpp>
#include <boost/interprocess/shared_memory_object.hpp>
#include <boost/interprocess/streams/bufferstream.hpp>
#include <iostream>
namespace bip = boost::interprocess;
int main() {
bip::shared_memory_object smo(bip::open_or_create, "MySharedMemory", bip::read_write);
std::string str = "test data";
smo.truncate(10ull << 10); // 10 KiB
bip::mapped_region r(smo, bip::read_write);
bip::bufferstream stream(reinterpret_cast<char*>(r.get_address()), r.get_size());
if (stream << str)
std::cout << "Written";
}
¹ https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem
² Coliru 不支持共享内存