How to check if element is included in array with arduino?

Question

I have an array called numbers and I want to test if a certain number is included in that array.

void loop(){
  if (arrayIncludeElement(numbers, mynumber)){
    // Do something
  } else {
    // Do something else
  }
}

I have written a version of the arrayIncludeElement function, but it don't seem to work.

boolean arrayIncludeElement(array, element) {
  for (int i = 0; i < sizeof(array); i++) {
    if (array[i] == element) {
      return true;
    }
  }
  return false;
}

ERROR MESSAGES

sketch:5: error: 'array' was not declared in this scope
sketch:5: error: 'element' was not declared in this scope
sketch:5: error: initializer expression list treated as compound expression
sketch:3: error: redefinition of 'boolean arrayIncludeElement'
sketch:5: error: 'boolean arrayIncludeElement' previously defined here
sketch.ino: In function 'void loop()':
sketch:11: error: 'arrayIncludeElement' cannot be used as a function
sketch.ino: At global scope:
sketch:16: error: redefinition of 'boolean arrayIncludeElement'
sketch:5: error: 'boolean arrayIncludeElement' previously defined here
sketch:16: error: 'array' was not declared in this scope
sketch:16: error: 'element' was not declared in this scope

EDIT I updated the function to this.

boolean arrayIncludeElement(int array[], int element);
boolean arrayIncludeElement(int array[], int element) {
  for (int i = 0; i < (sizeof(array)/4); i++) {
    if (array[i] == element) {
      return true;
    }
  }
  return false;
}

But now I get this error

sketch.ino: In function 'void loop()':
sketch:8: error: 'arrayIncludeElement' was not declared in this scope
sketch:10: error: expected `;' before '}' token

Line 8 being this

if (arrayIncludeElement(numbers, 5){

and Line 10 being this

} else {

Shouldn't you specify the types for your function arguments? — Dmitri
– Dmitri, Commented Jan 13, 2015 at 8:40
Also, sizeof(array) wouldn't work in your function -- you need to pass the size separately. — Dmitri
– Dmitri, Commented Jan 13, 2015 at 8:43
The sizeof(array) works alright. The problem was clearly that I missed the datatypes for arguments to the function. — Emil Bonne Kristiansen
– Emil Bonne Kristiansen, Commented Jan 13, 2015 at 8:48
No, the sizeof(array) gives the number of bytes in a pointer to int (or whatever your array is of). It gives something, but not the right thing. — Dmitri
– Dmitri, Commented Jan 13, 2015 at 8:49
Okay. But I don't know the length of my array, so is there another approach? — Emil Bonne Kristiansen
– Emil Bonne Kristiansen, Commented Jan 13, 2015 at 8:53

Karthikeyan.R.S · Accepted Answer · 2015-01-13 09:16:48Z

1

You are not mentioning the data type of that arguments. You have to mention the arguments. Then you have to declare the function prototype.

 boolean arrayIncludeElement(int array[], int element);
 boolean arrayIncludeElement(int array[], int element) {
 for (int i = 0; i < max; i++) {
      if (array[i] == element) {
          return true;
      }
    }
  return false;
 }

Edit: As like Dmitri said, pass the no of elements in the array as a function argument. Don't use the sizeof.

Then declartion of function, must be placed before where it is used. So in this place the function prototype before the function void loop().

edited Jan 13, 2015 at 9:16

answered Jan 13, 2015 at 8:42

Karthikeyan.R.S

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4 Comments

Karthikeyan.R.S Over a year ago

@EmilBonneKristiansen Ok no Problem.

Dmitri Over a year ago

Remember that inside the function, array looks like a pointer -- the type doesn't specify the number of elements. sizeof(array) will give the number of bytes in the pointer, not the number of bytes in the whole array. He's got to pass the size separately.

Emil Bonne Kristiansen Over a year ago

I thought it worked for a moment, but it seems there is a new error.

Emil Bonne Kristiansen Over a year ago

Okay, A friend of mine made it work. I feel like I am not really used to this kind of strict programming. I marked this answer as correct, thx everyone.

Community · Accepted Answer · 2017-05-23 11:50:56Z

0

It seems that this question is getting closer to How do I determine the size of my array in C?

BUT, in the general case, you do have to provide the length of the array or use some guard element (like null-terminated strings). Somewhere up the call chain you must know the size of the array, don't you?

A small example to show why sizeof() may not be correct:

int size_test(char b[])
{
    char a[10];
    char foo[] = {'a', 'b', 'c', 'd', 'e'};
    char *bar = malloc(20);
    printf("sizeof a = %ld, sizeof b = %ld, sizeof foo = %ld, sizeof bar = %ld\n", sizeof(a), sizeof(b), sizeof(foo), sizeof(bar));
}
int main()
{
    char b[12];
    size_test(b);
}

which, on my (64 bit) machine, prints

sizeof a = 10, sizeof b = 8, sizeof foo = 5, sizeof bar = 8

edited May 23, 2017 at 11:50

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answered Jan 13, 2015 at 9:15

drRobertz

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