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, Jan 13 '15 at 08:40
Also, `sizeof(array)` wouldn't work in your function -- you need to pass the size separately. — Dmitri, Jan 13 '15 at 08:43
The `sizeof(array)` works alright. The problem was clearly that I missed the datatypes for arguments to the function. — Emil Bonne Kristiansen, Jan 13 '15 at 08:48
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, Jan 13 '15 at 08:49
Okay. But I don't know the length of my array, so is there another approach? — Emil Bonne Kristiansen, Jan 13 '15 at 08:53
You'll either have to keep track of the array size, or mark the end with a reserved value (one that you don't consider valid in your array). — Dmitri, Jan 13 '15 at 09:01

Karthikeyan.R.S · Answer 1 · 2015-01-13T09:16:48.493

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 '15 at 09:16

answered Jan 13 '15 at 08:42

Karthikeyan.R.S

3,991
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@EmilBonneKristiansen Ok no Problem. – Karthikeyan.R.S Jan 13 '15 at 08:47
1

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. – Dmitri Jan 13 '15 at 08:54
I thought it worked for a moment, but it seems there is a new error. – Emil Bonne Kristiansen Jan 13 '15 at 09:00
1

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. – Emil Bonne Kristiansen Jan 13 '15 at 09:20

score 0 · Answer 2 · edited May 23 '17 at 11:50

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

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

2 Answers2