Passing array to function

[urbanspaceman]

Hi, i'm stuck with a (maybe) simple problem

i have a sketch like this (simplified version)

void myfunction(int lenght, uint16_t x, uint16_t y, uint16_t s, char t_name, uint16_t color, int myarray[]) {

  for (int i = 0; i < 16; i++) {
             if(myarray[i] == 1){
              //do something
             }else{
                if (lenght < i + 1) {
                  //do another thing
                } else {
                  //or another
                }
              }
  }
}

  void setup() {
    // put your setup code here, to run once:


  }

  void loop() {
    // put your main code here, to run repeatedly:
    int my_array = {1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0};


   Myfunction(14, 30, 60, 18, 'A', ILI9341_ORANGE, my_array);


  }

i have a problem passing the array to the function

 

[el_supremo]

    int my_array = {1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0};

That is not an array. This is:

    int my_array[16] = {1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0};

It also helps to spell the name of the function the same way. Myfunction and myfunction aren't the same.

Pete

 

[urbanspaceman]

Thank you El_supremo!
Now works!

 

[WMXZ]

If you don't know (or are too lacy to count) you could do

int array[]  = {1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0};

and then determine the size by

int n = sizeof(array)/sizeof(array[0]);

As you typically pass your array by pointer, you should pass also the length of the array

myFunction(array,n);

Edit: as it seems that sizeof of an operator and not a function

int n = sizeof array /sizeof array[0] ;

is sufficient (or more appropriate?)

 

[urbanspaceman]

works perfectly, thanks WMXZ