Do it on an int or uint type of data! What happens to the operation

Question

I'm having an issue with the ESP32-IDF auto-generated code, where the following code is used to flip the pin level by default in an example code of a flashing light

static uint8_t led_state = 0;

...

led_state = !led_state;  toggle the pin state

how to understand the '!' operation here

I write a small program like this

#include <stdio.h>
#include <stdint.h>

int main(void)
{
    uint8_t i = 2;
    printf("i = %d\n", i);
    i = !i;
    printf("i toggled\n");
    printf("i = %d\n", i);
    i = !i;
    printf("i toggled\n");
    printf("i = %d\n", i);
    return 0;
}

and the output is

i = 2
i toggled
i = 0
i toggled
i = 1

Answer 1

! is the logical NOT operator.

• If it's operand isn't equal to zero (i.e. if it's a true value), it returns 0 (a false value).
• If it's operand is equal to zero (i.e. if it's a false value), it returns 1 (a true value).

C logical operators

---

If you need to set a specific bit, you can use a shift.

led_state = !led_state;            // Toggle the state.
uint8_t bitmask = led_state << 1;  // 1 or 0 => 2 or 0.

If you want to toggle a specific bit, you can use an exclusive or.

bitmask ^= 1 << 2;                 // Toggle the second bit directly.