passing arrays by reference

Forum 2005-2010 (read only) Software Syntax & Programs
1

I am new to Arduino, but come from a c++ back ground in microsofts VS. I am trying to pass an array into a function by reference, so that more than one function can operare on the same array. An example of what I'm trying to do:

void function(int &);

void setup(){
Serial.begin(9600);

};

void loop(){
int counter[10];
funtion(&cunter[]);
for(int i=0;i<10;i++){
Serial.println(counter*);*

  • }*
    };
    void function(int &functionary[]){
  • for(int i=0;i<10;i++){*
    funtionary_=(i10);_
    _     }_
    _    }_
    _    [/quote]*_
    At least in VS. this is how I would do it, however arduino returns an error stating 'declaration of 'ary' as array of references'. I'm not really sure what to do as a work around. So far I have concluded that the prototype is needed to fix a bug in arduino's auto-prototyping, and I have found several postings saying you can not return an array from a function. Anything will help.
    rigg
2

Arrays (which actually are pointers) pass by reference by default.

void function(int arr[]);

void setup(){
  Serial.begin(9600);
  
}

void loop(){
  int counter[10];
  funtion(cunter);
  for(int i=0;i<10;i++){
    Serial.println(counter[i]);
  }
}

void function(int arr[]){
  for(int i=0;i<10;i++){
    arr[i]=(i*10);
  }
}

:slight_smile:

3

I corrected the spelling, the syntax and removed the extraneous semicolons:

const int arraySize = 10;
void setup(){
  Serial.begin(9600);
}

void loop(){
  int counter[arraySize];
  function(counter);
  for(int i=0;i<arraySize;i++){
    Serial.println(counter[i]);
  }
}

void function(int *functionary){
  for(int i=0; i < arraySize; ++i){
    functionary[i]=(i*10);
  }
}
4

Thank you Groove i know there are some spelling mistakes in the code. This is just a quick example of my code to clearly demonstrate the problem i was having.

AlphaBata- do the arrays pass by reference because of arduino's auto-prototype thing. Because I tryed just running code very similer to how you have it writen and did not get anything to print to the serial line. Should I leave off the prototype?

Thank you for your quick responce.

rigg

5

Because arrays are pointers, you're passing arrays by 'reference'.

You'll have to excuse my code, I was too quick on the Post button. It should be OK now.

int x[]; and int *x; are basically the exact same.

C++ can never pass an array by value, because of the nature of how arrays work.
The variable is just a pointer to the address of the first entry.

int x[10];

x[2] is the same as *(x+2*sizeof(int))

03:00
/me got to sleep! See you :slight_smile:

6

pass an array into a function by reference

In C(++), that isn't necessary. Arrays are always passed by reference.

At least in VS. this is how I would do it

No, you wouldn't. This...

  function(&counter[]);

...produces this...

error C2059: syntax error : ']'

...in Visual Studio 2008.

If you change to a prototype with a named argument this is the result...

void function(int &functionary[]);
error C2234: 'functionary' : arrays of references are illegal

Try dropping the reference...

void function(int functionary[]);

void setup(){
  Serial.begin(9600);
  
};

void loop(){
  int counter[10];
  function(counter);
  for(int i=0;i<10;i++){
    Serial.println(counter[i]);
  }
};

void function(int functionary[]){
  for(int i=0;i<10;i++){
    functionary[i]=(i*10);
  }
}
7

And we arrive back to reply #1 :slight_smile:

rigg: Have you solved the problem? Got some code that does what it's supposed to do?

8

Yes I beleive all the input has helped me solve my problem. It is going to be a few days before I can run the code to a microcontroller, however all the compiling errors have gone away. I will let every one know how it turns out in a couple of days.

Thank you to everybody how took time to help out, this is my first time asking for help in a forum and I am very thankfull with the quick responses.

rigg

9

AlphaBeta,

If you have an array of, for instance, intsint x[10];And an array reference likei = x[2];Then the pointer equivalent would bei = *(x + 2);There's no need to multiply by sizeof(int) -- C does this for you automatically.

Regards,

-Mike

10

AIf you have an array of, for instance, intsint x[10];And an array reference likei = x[2];Then the pointer equivalent would bei = *(x + 2);There's no need to multiply by sizeof(int) -- C does this for you automatically.

Ah. True that.

I just came from a C code dealing with void pointers.
:-[