CODEWITHSUNDEEP
c++arduinoribbon-control
Charles Julien
Charles Julien

Reputation: 465

delay function on Arduino doesn't behave as expected

I am working with an addressable ribbon with Arduino. The point is to light up different parts of my ribbon at different moment. To do this, I used the function delay as below:

void un_a()  //first third of ribbon length
{      
  for (uint16_t i = 0; i < N; i++) {
  strip.setPixelColor(i, strip.Color(100,255,100));
  }
  strip.show();
}

void deux_a()    //second third of ribbon length
{
  for (uint16_t i = N; i < 2*N; i++) {
    strip.setPixelColor(i, strip.Color(100,255,100));
  }
  strip.show();
}

void trois_a()    //last third of ribbon length
{
  for (uint16_t i = 2*N; i < 3*N; i++) {
    strip.setPixelColor(i, strip.Color(100,255,100));
  }
  strip.show();
}

void wave(){
  void un_a();
  delay(2000);
  void deux_a();
  delay(2000);
  void trois_a();
}

So when wave() is called, the expected behavior is:

In reality, it just blocks and lights a part of the 1st third.
I went around again and again, I don't see what I am missing. Any clue?

Upvotes: 0

Views: 74

Answers (1)

KamilCuk
KamilCuk

Reputation: 141533

void un_a();

This is a function declaration. It tells, that such symbol un_a exists, and it is a function of type void (*)().
If you want to call a function, you use a expresssion statement. Notice it does not have a returned type on the beginning as in a declaration:

int a; // declaration
a = 1; // statement
un_a(); // statement, this executed the un_a function
1 + 1; // another statement, this adds 1 + 1
int func(int b); // declaration, this does nothing, just the compiler knows that a function `func` exists
int (*(*func2)(int a, int (*(*)(int arr[a]))[a]))[5]; // another declaration
(void)func2(5, (int (* (*)(int *))[5])0); // statement

Try calling the functions:

void wave(){
  un_a();
  delay(2000);
  deux_a();
  delay(2000);
  trois_a();
}

Upvotes: 3

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