Question

1. In a C program for a microcontroller, portb has been setup as all outputs and is assigned the hexadecimal value 0xC6. Determine the voltage levels on the pins of portb.

2. In the context of explaining the difference between the logical NOT operator(!) and the bitwise NOT operator(~) say what value portb will end up with in each of the following cases:-

1. portb = ~ (0x45);
2. portb = !(0x45);

3. Determine the value on the portb pins after the following C code statement is executed, if portb had an initial value of 0x52:-

portb = portb >> 2;

4. In the context of explaining the difference between the logical AND operator(&&) and the bitwise AND operator(&) say what value portb will end up with in each of the following cases, if the initial value of portb is 0x77:-

1. portb = portb & 0x0F;
2. portb = portb && 0x0F;

Answer 1

Answer :- i) Value is 0xC6 = 1100_0110 so

pin0 = 0 V
pin1 = Vcc
pin2 = Vcc
pin3 = 0 V

pin4 = 0 V
pin5 = 0 V
pin6 = Vcc
pin7 = Vcc

Answer :- ii)a) ~(0x45) = ~(0100_0101) = 1011_1010 = 0xBA. So portb = 0xBA
b) !(0x45) = 0x00 since logical NOT of non-zero is zero. So portb = 0x00.

Answer :- iii) (0x52 >> 2) = (0101_0010 >> 2) = 0001_0100 = 0x14. Thus portb = 0x14.

Answer :- iv)a) (0x77 & 0x0F) = (0111_0111) & (0000_1111) = 0000_0111 = 0x07. Thus portb = 0x07.
b) 0x77 && 0x0F = 0x01, logical operators give result either 0 or 1. Since here both operands are non-zero so logical AND of them will be logic 1.