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1

I'm kind of new to c++ and my question might have a pretty easy solution, but I couldn't figure it out by myself.

Let's say I have two byte arrays a and b. Each of them contains six bytes. Now I want to introduce a new array c which should contain a and b.

This is how I tried it:

byte a[] = {B11111111, B10010000, B10011000, B10010100, B10010010, B11110001};
byte b[] = {B11111111, B10000001, B10000001, B10000001, B10000001, B11111111};

byte c[2][6] = {{a},{b}};

The compiler gives me the following error:

invalid conversion from 'byte' to 'byte'
share|improve this question
    
Never mind that you're confusing addition and multiplication (or maybe I should say unions and products), but the solution I'm envisaging uses several traits to extract array sizes and templates of variable numbers of index packs to initialize a single union array from arbitrary constituents. I think that might be a bit more involved than you were looking for. – Kerrek SB Jul 13 '13 at 0:19
    
Your question embodies a contradiction in terms. A byte array is an array of bytes. An array of byte arrays is an array of arrays. – EJP Jul 13 '13 at 0:20
    
So, the question should be "how to create an array of byte arrays". It seems I can't edit questions jet. But there must be a smple solution to this anyway. – user1210456 Jul 13 '13 at 0:23
    
in c# it works like I tried it : stackoverflow.com/questions/549399/… how can I do this in c++? – user1210456 Jul 13 '13 at 0:28

3 Answers 3

up vote 2 down vote accepted

You could do it as follows:

byte a[] = {B11111111, B10010000, B10011000, B10010100, B10010010, B11110001};
byte b[] = {B11111111, B10000001, B10000001, B10000001, B10000001, B11111111};

byte* c[2] = {a,b};

But it would be cleaner to just do a multidimensional array directly:

byte c[2][6] = {
  {B11111111, B10010000, B10011000, B10010100, B10010010, B11110001},
  {B11111111, B10000001, B10000001, B10000001, B10000001, B11111111}
};
share|improve this answer
    
This works. Can you explain why it is cleaner to do it directly and what exactly does byte*. I have to create the multidimensional array later in the code because a and b represent bitmaps for letters and c is a word. Which word it is, should be variable. – user1210456 Jul 13 '13 at 0:52
    
byte* c[2] is an array holding 2 byte pointers. A byte[] array can be degraded into a byte* pointer. So c is merely pointing at the starting memory addresses of a and b, rather than making copies of them. – Remy Lebeau Jul 13 '13 at 1:35
    
It is cleaner if you're simply declaring these at the top of your code somewhere, since it's all in one, simple object without any unnecessary intermediary objects (e.g. a and b). However, if you have a and b anyway and just want to hold them together, then the first solution makes more sense. As Remy explained, byte * c[2] is just an array of byte*, so each entry in c[] is a pointer - and hence an array - in itself. You can think of byte a[] as being the same as byte *a. – Skipper E Jul 13 '13 at 17:17
up vote 5 down vote

Raw arrays are a bit annoying. Use std::array instead:

using std::array;
array<byte,6> a = {B11111111, B10010000, B10011000, B10010100, B10010010, B11110001};
array<byte,6> b = {B11111111, B10000001, B10000001, B10000001, B10000001, B11111111};
array<array<byte,6>,2> c = {a, b};

std::array was introduced in

share|improve this answer
    
This sounds like a good solution but I'm wirting the code for an arduino and I get the following error: expected constructor, destructor, or type conversion before '<' token. – user1210456 Jul 13 '13 at 1:24
    
This solution is also a coorect suggestion. Arduino doesn't support std::array and std::vector. I marked the pointer solution as the right one because it's a good solution and it works for me. Anyway, using std::array might be a better soloution in some other cases. I will vote +1 as soon as my reputation allows it. – user1210456 Jul 13 '13 at 13:54
up vote 0 down vote

you have to do a for loop to copy the 2 arrays into the third,

for (int i = 0; i < 6; i++)
{
   c[0][i] = a[i];
   c[1][i] = b[i];
}
share|improve this answer

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