const char *到std :: basic_iostream [英] const char * to std::basic_iostream

问题描述

我有一个指向 const * char 缓冲区的指针及其长度，并尝试使用API​​（在这种情况下， AWS S3 C ++上传请求），它接受以下类型的对象：

I have a pointer to a const *char buffer as well as it's length, and am trying to use an API (in this case, the AWS S3 C++ upload request) that accepts an object of type:

std::basic_iostream <char, std::char_traits <char>>

是否有一种简单的标准C ++ 11方法将缓冲区转换为兼容的流，最好不要实际复制了内存？

Is there a simple standard C++11 way to convert my buffer into a compatible stream, preferably without actually copying over the memory?

推荐答案

由于Igor的评论，这似乎行得通：

Thanks to Igor's comment, this seems to work:

func(const * char buffer, std::size_t buffersize)
{     
    auto sstream = std::make_shared<std::stringstream>();
    sstream->write(buffer, buffersize);
    ...
    uploadRequest.SetBody(sstream);     
    ....

这篇关于const char *到std :: basic_iostream的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！