initialize this buffer uint8*

Question

I have this typedef

typedef unsigned char uint8;

and this variable

public : uint8* bufferOfExchange;

how could I initialize this buffer?

bufferOfExchange = ???

The answer to this depends on the value that you want it to have once initialised. — Mankarse, Nov 24 '11 at 15:44
Maybe `new uint8` or `new uint8[size]` or maybe U don't really need pointer... — Hauleth, Nov 24 '11 at 15:46
You could initialize it to null bufferOfExchange = 0; or you could allocate memory bufferOfExchange = (uint8*)malloc(1024); and so on — Cyclonecode, Nov 24 '11 at 15:46
I dlike to have something like byte[] from c# =). So c#(byte[])->c++(uint8* buffer) — curiousity, Nov 24 '11 at 15:46
@curiousity- In that case you should be using `std::vector` instead of `uint8*`. — Mankarse, Nov 24 '11 at 15:47

Nawaz · Answer 1 · 2011-11-24T15:51:01.657

2

Like this:

bufferOfExchange = new uint8[bufferSize]; //bufferSize is size_t type. 

//or
bufferOfExchange = otherBuffer; //otherBuffer is of same type

What else do you think?

Better choice would be to use std::vector<uint8> instead of uint8*:

std::vector<uint8> bufferOfExchange;

Now, read some good book to know how to use std::vector.

edited Nov 24 '11 at 15:51

answered Nov 24 '11 at 15:46

Nawaz

I dlike to have something like byte[] from c# =). So c#(byte[])->c++(uint8* buffer) is it possible? – curiousity Nov 24 '11 at 15:49
@curiousity: I suggested you to use `std::vector`. – Nawaz Nov 24 '11 at 15:51

score 1 · Answer 2 · answered Nov 24 '11 at 15:46

Well you don't have a buffer, only an uninitialized pointer. You can create a buffer with new like this:

bufferOfExchange = new uint8[10];

(10 is an arbitrary choice - use the buffer size you need.)

For real code however, you would probably want std::vector<uint8>.

2 Answers2