Reading raw byte array into std::string

Question

I've been wondering about the following issue: assume I have a C style function that reads raw data into a buffer

int recv_n(int handle, void* buf, size_t len);

Can I read the data directly into an std:string or stringstream without allocating any temporal buffers? For example,

std::string s(100, '\0');
recv_n(handle, s.data(), 100);

I guess this solution has an undefined outcome, because, afaik, string::c_str and string::data might return a temporal location and not necessarily return the pointer to the real place in the memory, used by the object to store the data.

Any ideas?

chrisaycock · Accepted Answer · 2010-12-23 17:04:01Z

12

Why not use a vector<char> instead of a string? That way you can do:

vector<char> v(100, '\0');
recv_n(handle, &v[0], 100);

This seems more idiomatic to me, especially since you aren't using it as a string (you say it's raw data).

answered Dec 23, 2010 at 17:04

chrisaycock

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3 Comments

Chinasaur Over a year ago

C++11: recv_n(handle, v.data(), 100);

Mooing Duck Over a year ago

@Chinasaur: I think a cast is required there, because of the const.

Chinasaur Over a year ago

Confusingly, vector::data() is not const but string::data() is.

Chinasaur · Accepted Answer · 2014-03-24 20:58:18Z

11

Yes, after C++11.

But you cant use s.data() as it returns a char const*

Try:

std::string s(100, '\0');
recv_n(handle, &s[0], 100);

Depending on situation, I may have chosen a std::vector<char> especially for raw data (though it would all depend on usage of the data in your application).

edited Mar 24, 2014 at 20:58

Chinasaur

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answered Dec 23, 2010 at 17:02

Loki Astari

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5 Comments

FireAphis Over a year ago

I think this might not work because s[0] is the same as s.data()[0], which might point to a temporal location.

Loki Astari Over a year ago

@FireAphis: No. s[0] is not the same as s.data()[0]. The difference is that if the string uses a copy on write implementation it has to make sure that the string is unique before it returns a reference. That is why s[0] returns a reference to a char and not a reference to a const char. The reason data() returns a const char* is that the string data may be being shared with multiple instances of string. Thus when you do &s[0] you are guaranteed to have a pointer to a contiguous unshared string.

Nim Over a year ago

I don't think this is a safe approach either, primarily as I believe that std::string doesn't make any guarantees about storage being contiguous - ofcourse for such a small block, it probably is, however there is no guarantee (unless this is C++x0), @chrisaycock's approach is safer.

Loki Astari Over a year ago

@Nim: You are correct. I don't believe the C++03 standard guarantees it. But I believe that C++0x does (because they wanted to be able to pass mutatable string to C-Interfaces) and as part of the analysis they did was to look at current implementations and found that no current implementations of the STL did not use contiguous space for &s[0].

Mooing Duck Over a year ago

It is indeed guaranteed in C++11. In C++03 it wasn't guaranteed, but it was impracticable to implement a string any other way, so yeah, everyone made it contiguous.

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