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[Reply #15 on:]

I see now - I had misread what the bitWrite was doing. The array only holds the pin numbers, not the actual result.

[Reply #16 on:]

Well you didn't read the sticky or your code would be in code tags. Yes, it's a big thing. A really big thing.

Thanks sir Nick. I was talking about the 5-hour thing and about my "UP" reply. noted.

If you have 4 switches, one option is to convert those 4 inputs into a number:

Code:

BCD_number = 0; // clear the variable to start
BCD_number = digitalRead(Din);  // result is 0 or 1
BCD_number = BCD_Number <<1 + digitalRead(Cin); // shift over 1 bit, and add in next bit
BCD_number = BCD_Number <<1 + digitalRead(Bin); // shift over 1 bit, and add in next bit
BCD_number = BCD_Number <<1 + digitalRead(Ain); // shift over 1 bit, and add in next bit
Serial.println (BCD_Number, DEC); // check the result

// Now use your display mapping:
shiftout(dataPin, clockPin, MSBFIRST, numbers[BCD_number] ); // lookup the mapping in the array and send it to the shift register

 

Thank you so much sir! By the way, I just have to
add this:
BCD_number = BCD_Number <<1 + digitalRead(Din); // shift over 1 bit, and add in next bit

and then copy-paste all these lines to the loop function, right?

Also, is there anything else I need to modify in the codes to display two hex digits in two 7 segment display? How am I suppose to connect the second shift register to the first one?

[Reply #17 on:]

Yes, read the switches, make the result into a number.
Use the number to look up the character you want to send out.
Shift registers connect like this.
I don't know if your mapping has 1 = on for a segment, or if your display uses common cathode or common anode parts.
With common cathode, 1 = on for a segment. WIth common anode, 0 = on for a segment.
Don't forget the caps on the +5V pin for reliable operation.