Array question

Projects Programming
1

Hello, I'm an old newbie to c++ , so sorry if question sounds stupid :confused:

I'm trying to save some ram by reusing an array

This does not compile :

      // declaring the array
       byte myChar[8] = {
		B00000,
		B10001,
		B00000,
		B00000,
		B10001,
		B01110,
		B00000,
		B00000
	};

	lcd.createChar(0, myChar);


        //attempt to refill that same array , but it fails
	myChar[8] = {
		B00000,
		B10001,
		B00100,
		B00000,
		B10001,
		B01110,
		B00100,
		B00000
	};
	lcd.createChar(1, myChar);

Since , - I believe - every time I write " byte araynam[8] {} " I lose 8 byte in RAM I toucht why not resuse 1 array and fill it up again esaily just like I declared it.
Of course I could go through every element in the array by using its indexnr between the [] and do an assign but that does not look fine.

Anyone ?

Sorry for bad english, I speak dutch

2

You are right. You cannot define the contents of the array like that a second time. You need to set each element individually, perhaps by copying from another array element by element or using memcpy() to copy the whole of another array.

Example

char array1[] = {"test"};
char array2[] = {"0123"};

void setup()
{
  Serial.begin(115200);
  Serial.print("array1 ");
  Serial.println(array1);
  Serial.print("array2 ");
  Serial.println(array2);
  memcpy(array2, array1, sizeof(array2));
  Serial.print("array2 ");
  Serial.println(array2);
}

void loop()
{
}
3

Hey UKHeliBob, I was afraid for this that there would not be an "proper" second chance to reassign the values.

I'll try to do a memcpy from a const array, see if that works. Thanks for the tip.

Maybe even better would be that I could rewrite the lcd.createchar function in
such way that I could enter the 8 elements in the function call, but in C++ , I have
no doubt how to do that. Got lots of learning to do

Grtz

4

Be aware that if you define a character and display it then if it is still on the display when you redefine it then the character already displayed will change as well.