将uint8_t *缓冲区上传到AWS S3，而无需通过文件系统 [英] Upload uint8_t* buffer to AWS S3 without going via filesystem

问题描述

免责声明: 我不是c ++程序员，请从我自己身上救出来.

Disclaimers: I am not a c++ programmer, please save me from myself.

我正在尝试使用AWS开发工具包在c ++中创建一个PutObjectRequest.

I'm trying to create a PutObjectRequest in c++ using the AWS SDK.

我有一个'uint8_t *'(在Java领域，从这里开始，我称其为byte []，我相信在c + +上这是一个缓冲区)，我需要将其放入Aws :: IOStream不知何故.

I have a 'uint8_t*' (in Java-land from whence I hail we call this a byte[], I believe on planet c++ this is a buffer), and I need to get it into an Aws::IOStream somehow.

所有示例都显示了直接来自文件系统的数据.

All of the examples show the data coming directly from the filesystem.

我看过几个类似(但不是真的)的问题，它们的答案指向另一个称为Boost的第三方库，但是确定这是一个常见用例吗?为什么我需要第三方库来做一些可以使用AWS开发工具包完成的事情?:

I've seen a couple of similar-ish (but not really) questions with answers that point to another third party library called Boost, but surely this is a common usecase? Why would I need a third party library to do something that should just be possible using the AWS SDK?:

我有数据，我想将其放在S3上.不，它不在文件系统中，是的，我在内存中创建了它."

uint8_t* buf; //<-- How do I get this...
...
Aws::S3::Model::PutObjectRequest object_request;
object_request.WithBucket(output_bucket).WithKey(key_name);

object_request.SetBody(data); //<-- ...into here

我真的很感谢这里的任何帮助或指点(无双关语).

I really appreciate any help or pointers (no pun intended) here.

更新
我已经尝试了评论中的所有内容，并且:

Update
I've tried everything in the comments, and this:

std::shared_ptr<Aws::IOStream> objectStream = Aws::MakeShared<Aws::StringStream>("PutObjectInputStream");        
*objectStream << data;
objectStream->flush();
object_request.SetBody(objectStream);

和这个:

std::shared_ptr<Aws::IOStream> objectStream = Aws::MakeShared<Aws::StringStream>("PutObjectInputStream");        
std::istringstream is((char*) data);
*objectStream << is.rdbuf();
objectStream->flush();
object_request.SetBody(objectStream);

可以编译，但每个只能上传2个字节的数据.

which compile, but each only uploads 2 bytes of data.

我尝试过的其他无法编译的内容是:

Other thing I've tried that don't compile are:

auto input_data = Aws::MakeShared<Aws::IOStream>("PutObjectInputStream", std::istringstream((char*) data), std::ios_base::in | std::ios_base::binary);
object_request.SetBody(input_data);

和

object_request.SetBody(std::make_shared<std::istringstream>( std::istringstream( (char*) spn ) ));

这些将在S3上创建对象，但具有0个字节:

and these ones creates the object on S3, but with 0 bytes:

    std::shared_ptr<Aws::IOStream> objectStream = Aws::MakeShared<Aws::StringStream>("PutObjectInputStream");
    objectStream->rdbuf()->pubsetbuf(static_cast<char*>(reinterpret_cast<char*>(data)), length);
    objectStream->rdbuf()->pubseekpos(length);
    objectStream->seekg(0);
    object_request.SetBody(objectStream);


    std::shared_ptr<Aws::IOStream> objectStream = Aws::MakeShared<Aws::StringStream>("PutObjectInputStream");
    objectStream->rdbuf()->pubsetbuf(reinterpret_cast<char*>(data), length);
    objectStream->rdbuf()->pubseekpos(length);
    objectStream->seekg(0);
    object_request.SetBody(objectStream);

推荐答案

在小时被黑客入侵之后，答案是:

And after hours of hacking, here is the answer:

    Aws::S3::Model::PutObjectRequest object_request;
    object_request.WithBucket(output_bucket).WithKey(key_name);
    auto data = Aws::MakeShared<Aws::StringStream>("PutObjectInputStream", std::stringstream::in | std::stringstream::out | std::stringstream::binary);
    data->write(reinterpret_cast<char*>(buffer), length);
    object_request.SetBody(data);

感谢 Ben Voigt 指出，当您make_shared(或本例中的MakeShared)时， )，那么您实际上并没有将数据传递给它.您只是告诉它要共享什么T.

Thanks to Ben Voigt for pointing out that when you make_shared (or MakeShared in my case), you're not actually passing it the data at that point. You're just telling it what T you're making shared.

也由nbubis回答了以下问题: const char *到std :: basic_iostream

Also helped by nbubis' answer to this question: const char * to std::basic_iostream

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