Convert uint8_t* to std::string in C++?

Question

Possible Duplicate:
How to Convert Byte* to std::string in C++?

I'm on an embedded device and try to receive a message. This message is given by a const uint8_t* data and its length size_t len.

Now I need a std::string to output my data.

Looks like a duplicate of this: http://stackoverflow.com/q/1673445/50079 — Jon, Dec 22 '10 at 11:57
Voting to reopen, since this is from `uint8_t` and the other is `char` which might be different on insane devices. — Ciro Santilli Путлер Капут 六四事, Nov 19 '18 at 11:45

score 15 · Accepted Answer · answered Dec 22 '10 at 12:07

15

If you don't want to convert the encoding, this will work:

std::string s( data, data+len );

If you want to convert UTF-8 into whatever system encoding is used by your platform, you need to use some platform-specific means.

answered Dec 22 '10 at 12:07

sbi

score 1 · Answer 2 · answered Dec 22 '10 at 11:57

1

Is your uint8* string null-terminated? If so, you can just do:

std::string mystring(data);

answered Dec 22 '10 at 11:57

when there's a length, the data is probably not null-terminated – Milan Dec 22 '10 at 12:08
And it's probably better to explicitly specify it anyway; all that's going to do is search for the length again anyway. – GManNickG Dec 22 '10 at 12:58
Also you are assuming that there is no '\0' character inbedded in the string. The std::string does not put any special emphasis on the '\0' character. – Martin York Dec 22 '10 at 19:56

Flinsch · Answer 3 · 2010-12-22T12:12:30.503

0

For some case sizeof(uint8) != sizeof(char), you could do:

std::string output( len, 0 );
for ( size_t i = 0; i < len; ++i )
  output[ i ] = static_cast<char>( data[ i ] );

edited Dec 22 '10 at 12:12

answered Dec 22 '10 at 12:00

Flinsch

3

`sizeof(uint8)!=sizeof(char)` is impossible. `sizeof` is in units of `char`, so if `char` is larger than 8 bits, there is no way to have an 8-bit type, i.e. `uint8` cannot exist. – R.. GitHub STOP HELPING ICE Dec 22 '10 at 16:13

3 Answers3