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The following code sets an edge-triggered interrupt to be which one of the following?

```c
GPIO_PORTF_IS_R &= ~0x20;
GPIO_PORTF_IBE_R &= ~0x20;
GPIO_PORTF_IEV_R &= ~0x20;
```

A. Rising Edge
B. Both Edge
C. None of the above
D. Falling Edge

Answer :

The code is specifically setting the interrupt to be triggered on a falling edge, indicating that the edge-triggered interrupt is a Falling Edge.

The code GPIO_PORTF_IEV_R &= -0x20; sets an edge-triggered interrupt to be a Falling Edge.

In this code, GPIO_PORTF_IEV_R is the register used to configure the interrupt event. By performing a bitwise AND operation with -0x20, which is equivalent to binary 11100000, the code clears the corresponding bit 5 while leaving the other bits unchanged. This means that only the falling edge of the input signal will trigger the interrupt.

Therefore, the code is specifically setting the interrupt to be triggered on a falling edge, indicating that the edge-triggered interrupt is a Falling Edge.

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https://brainly.com/question/32216925

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