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Question 1 0/4 pts 6.3 Which line of C code is a friendly way to set Port B bit 2 assuming this pin has already
Question 1 0/4 pts 6.3 Which line of C code is a friendly way to set Port B bit 2 assuming this pin has already been initialized as an output? GPIO_PORTB_DATA_R = 6x00; GPIO_PORTB_DATA_R = 0x02; GPIO_PORTB_DATA_R = 0x04; GPIO_PORTB_DATA_R I= 0x02; GPIO_PORTB_DATA_R I- Ox4; GPIO_PORTB_DATA_R &- 0x02; GPIO_PORTB_DATA_R &- 6x64; GPIO_PORTB_DATA_R &= -0x02; GPIO_PORTB_DATA_R &= 0x04; GPIO PORTB_DATA_R = 0x00; GPIO PORTB_DATA_R&= --Ox04; GPIO PORTB DATA R = Ox02 Not quite... Bitwise OR operation is used to set a bit. So that's good. But, what value should the mask be? GPIO PORTB_DATA_R = 0x02; GPIO PORTB DATA R&= Ox02; GPIO PORTB DATA R&=0x02; GPIO PORTB DATA R = Ox04; GPIO PORTB_DATA_R & Ox04; GPIO_PORTB_DATA_R I = 0x04: With practice, you will notice and use common expressions and patterns when setting, clearing and testing bits of a register. Question 2 0 / 4 pts 6.4 Which line of C code is a friendly way to clear Port B bit 2 assuming this pin has already been initialized as an output? GPIO_PORTB_DATA_R - 0x00; GPIO_PORTB_DATA_R - Ox02; GPIO_PORTB_DATA_R - 0x04; GPIO_PORTB_DATA_R I = 0x02; GPIO_PORTB_DATA_R I = 0x04; GPIO_PORTB_DATA_R &- 0x02; GPIO_PORTB_DATA_R &= 0x04; GPIO_PORTB_DATA_R &= ~@x02; GPIO_PORTB_DATA_R &= -x04; GPIO_PORTB_DATA_R = 0x00; GPIO PORTB_DATA_R = 0x04; GPIO PORTB DATA_R = 0x02; GPIO PORTB_DATA_R = 0x04; GPIO PORTB_DATA_R = 0x02; GPIO PORTB DATA R&=-Ox04: GPIO PORTB_DATA_R & Ox02: GPIO_PORTB_DATA_R &=0x04; GPIO PORTB DATA R&= --Ox02; Not quite... Bitwise AND operation is used to clear a bit. So that's good. But, what value should the mask be? With practice, you will notice and use common expressions and patterns when setting, clearing and testing bits of a register
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