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I am coding on the Arduino platform and I am trying to write something that will concatenate/append byte arrays in C.

byte a[] = {a1, ..., an};
byte b[] = {b1, ..., bm};

byte c[] = a + b; // equivalent to {a1, ..., an, b1, ..., bm}

What is the best way to get the above result?

I tried searching online, however I have not had much luck. I saw another answer on SO highlighting the steps needed in order to do this however I could not follow them. They also say that there are libraries that deal with this kind of thing, however as I am on Arduino I am unsure whether or not these are totally available to me.

I understand there needs to be some sort of memory manipulation in order for this to work however I am new to these kinds of low level manipulations so they do not make too much sense to me. I have experience in higher languages (C#, Java and some C++).

I should also add: Can the same technique work for:

byte a[] = {a1, ..., an};
byte b[] = {b1, ..., bm};

a = a + b
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  • Are the sizes fixed? Commented Feb 21, 2015 at 22:10
  • assuming byte is a char sized equivalent, byte c[sizeof(a) + sizeof(b]; and two memcpy calls would do it if that really is how a and b are declared. Commented Feb 21, 2015 at 22:11
  • @WhozCraig Yes. a and b are declared like this. However, I have edited the question to contain the case for when b is appended to a as well. Commented Feb 21, 2015 at 22:15
  • @Alizter that isn't going to happen unless a is initially sized to accommodate its data and space for appended data And a memcpy-type operation is still involved sooner or later. Regardless, the syntax you're using is definitely not congruent with the C language. Commented Feb 21, 2015 at 22:17
  • @WhozCraig Yes. I am aware that it is definitely not C. I have written it in a pseudo style to better explain my trouble. Commented Feb 21, 2015 at 22:19

2 Answers 2

Reset to default
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There is no byte type in C. Unless it's some type definition, you could use unsigned char or some fixed type from <stdint.h> portably. Anyway, here is some solution:

#include <stdio.h>
#include <string.h>

int main(void) {
    unsigned char a[3+3] = {1, 2, 3}; // n+m
    unsigned char b[3]   = {4, 5, 6}; // m

    memcpy(a+3, b, 3); // a+n is destination, b is source and third argument is m

    for (int i = 0; i < 6; i++) {
        printf("%d\n", a[i]);
    }

    return 0;
}

Make sure that array a has room for at least n + m elements (here n = 3 and m = 3 as well) in order to avoid issues with array overflow (i.e. undefined behavior, that may crash yor program or even worse).

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5 Comments

How would you make sure that a has enough room?
@Alizter unsigned char a[3 + 3] has room for the initial a + the appended b obviously.
@iharob Oh! I though there was some size I was missing. Of course.
So if we were to talk about an array c, assuming a and b are full, we would need to make c the 'length' of a + b and then copy using memcpy() into c b then a. What I am unsure of though is the a+3 syntax you have used. What does that mean?
@Alizter: Oh, my cognition only registered a=a+b case. In order of c=a+b you need to copy a to c, then b to c+n. The c+n works in the same way as a+3. The a "decays" (or more formaly is converted) to address of first element in array. By + operator with integer you are performing pointer arithmetic. The a+3 is equivalent to &a[3] so I believe you get an idea now.
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If the sizes are known you can simply create a new byte array of that size and use simple loops to fill it.

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1 Comment

No, no loops need to be involved.

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