Why shiftOut works when an inexisting position of an array is sent out?

Projects Programming
1

Following code works fine, but I do not understand why.
The array contains 10 items, the index for the counter is 16; for counter > 10, which value of the array is getting to sent out?

byte seven_seg_digits_dec[] = { B11111100,  // = 0
                                B01100000,  // = 1
                                B11011010,  // = 2
                                B11110010,  // = 3
                                B01100110,  // = 4
                                B10110110,  // = 5
                                B10111110,  // = 6
                               B11100000,  // = 7
                                B11111110,  // = 8
                                B11100110,   // = 9
                              };
void loop() {
  for (int bin = 0b0000; 16 > bin; bin++)
  {
   
 
      digitalWrite(latchPin1, LOW);
      shiftOut(dataPin1, clockPin1, LSBFIRST, seven_seg_digits_dec[bin]);
      digitalWrite(latchPin1, HIGH);
}
2

Some randomly stored bits in some memory address that is not being used by another thingie. Call it luck that things did not go poo!.

3

C++ does not check array boundaries. It is the responsibility of the programmer not to access beyond array boundaries. If you do so anyway, it's a coincidence what happens.

4

You cannot be shure about that. Usually you will get a conflict with another 'thingie'

5

So far the OP has got lucky on that point.

6

Yes, he's got lucky because he only reads from that array. If you also write to it ...

7

Which may be the next posting.

Help I made a 5 element array and when I write to element 6, my MCU crashes.

8

Even worse: :wink:
Help I made a 5 element array and when I write to element No. 5, my MCU crashes.

Ok, we should not hijack this topic. But neverless I think @homer_simpson-25 should read everything carfully ...