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Topic: passing arrays by reference (Read 831 times) previous topic - next topic

rigg

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:

Quote


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=(i*10);
  }
}



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

AlphaBeta

#1
Oct 23, 2009, 09:17 pm Last Edit: Oct 24, 2009, 02:49 am by AlphaBeta Reason: 1
Arrays (which actually are pointers) pass by reference by default.

Code: [Select]

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);
 }
}

:)

Groove

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

Code: [Select]
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);
 }
}
Per Arduino ad Astra

rigg

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

AlphaBeta

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
Code: [Select]
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];
Code: [Select]
x[2] is the same as
Code: [Select]
*(x+2*sizeof(int))

03:00
/me got to sleep! See you  :)

Coding Badly

#5
Oct 24, 2009, 03:36 am Last Edit: Oct 24, 2009, 03:37 am by bcook Reason: 1
Quote
pass an array into a function by reference


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

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


No, you wouldn't.  This...

Code: [Select]
 function(&counter[]);

...produces this...

Quote
error C2059: syntax error : ']'


...in Visual Studio 2008.


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

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



Try dropping the reference...

Code: [Select]
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);
 }
}

AlphaBeta

And we arrive back to reply #1  :)

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

rigg

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

Mike Murdock

AlphaBeta,

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

Regards,

-Mike

AlphaBeta

Quote
AIf you have an array of, for instance, ints
Code: [Select]
int x[10];And an array reference like
Code: [Select]
i = x[2];Then the pointer equivalent would be
Code: [Select]
i = *(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.
:-[

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