Study the following circuit:

Question

Suppose S1 and S2 are both not pressed down.

void setup() {
Serial.begin(9600);
pinMode(13, OUTPUT);
pinMode(2, INPUT);
pinMode(3, INPUT);
}

void loop() {
delay(2000);
int button1_val = !digitalRead(2); // button1_val = 1 when pressed, 0 when not pressed
int button2_val = !digitalRead(3); // button2_val = 1 when pressed, 0 when not pressed

if (button1_val == 1 && button2_val == 1) { // AND = && (both of them)
digitalWrite(13, 1); // turn on
}
else {
digitalWrite(13, 0); // turn off
}
}

Use the following codes to describe the state of the LED, and values of the variables (Your answer should either be 0, 1, 2, 3, 4, 5, 6, or 7)

CODE LED button1_val button2_val
0 OFF 0 0
1 OFF 0 1
2 OFF 1 0
3 OFF 1 1
4 ON 0 0
5 ON 0 1
6 ON 1 0
7 ON 1 1

Question

Study the following circuit:

Question

Suppose S1 and S2 are both not pressed down.

void setup() {
  Serial.begin(9600);
  pinMode(13, OUTPUT);
  pinMode(2, INPUT);
  pinMode(3, INPUT);
}

void loop() {
  delay(2000);
  int button1_val = !digitalRead(2); // button1_val = 1 when pressed, 0 when not pressed
  int button2_val = !digitalRead(3); // button2_val = 1 when pressed, 0 when not pressed
  
  if (button1_val == 1 && button2_val == 1) { // AND = && (both of them)
    digitalWrite(13, 1); // turn on
  }
  else {
    digitalWrite(13, 0); // turn off
  }
}

Use the following codes to describe the state of the LED, and values of the variables (Your answer should either be 0, 1, 2, 3, 4, 5, 6, or 7)

CODE	LED	button1_val	button2_val
0	OFF	0	0
1	OFF	0	1
2	OFF	1	0
3	OFF	1	1
4	ON	0	0
5	ON	0	1
6	ON	1	0
7	ON	1	1

BlackTom AI · Accepted Answer

Because the input pins are pulled up to 5V through the 10 kΩ resistors when their switches are open, digitalRead(2) and digitalRead(3) will each return HIGH (1). The code then applies the logical NOT operator (!) to those readings, so both button1_val and button2_val become 0. Since the if-statement requires both to be 1 (pressed) to turn the LED on, the LED remains off. This corresponds to code 0.

1) Relevant concepts
- Pull-up resistor: when a switch to ground is open, the input is held HIGH by the resistor.
- digitalRead returns HIGH (1) or LOW (0).
- The C-style ! operator inverts truth: !1 -> 0, !0 -> 1.
- The condition uses logical AND (&&): true only if both operands are 1.

2) Step-by-step reasoning
- S1 and S2 are both not pressed → both input nodes are pulled to 5V.
- digitalRead(2) = HIGH = 1.
- digitalRead(3) = HIGH = 1.
- button1_val = !digitalRead(2) = !1 = 0.
- button2_val = !digitalRead(3) = !1 = 0.
- if (button1_val == 1 && button2_val == 1) evaluates as (0 == 1 && 0 == 1) → (false && false) → false.
- else branch runs: digitalWrite(13, 0) → LED OFF.

3) Intermediate steps/calculations
- digitalRead(2) = 1
- !digitalRead(2) = !1 = 0 → button1_val = 0
- digitalRead(3) = 1
- !digitalRead(3) = !1 = 0 → button2_val = 0
- Condition: (0 == 1 && 0 == 1) → false → LED = 0

4) Why the correct answer is correct
The hardware produces HIGH when switches are open; the software negates that reading to produce 0 for each button variable. Because both variables are 0, the AND test fails and the LED is turned off. That matches the table row with LED OFF, button1_val = 0, button2_val = 0.

5) Why other options are incorrect (brief)
- Codes 1–3: also show LED OFF but require at least one button_val = 1; not possible here because neither switch is pressed.
- Codes 4–7: show LED ON and would require both button1_val and button2_val be 1 (both switches pressed or inputs read LOW so !0 = 1), which contradicts “both not pressed.”

6) Conclusion
The correct code is 0, matching the provided answer.

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