No: Pin 28 is:
Code:
28 EMC_32 SEMC_DATA10 FLEXPWM3_PWMB01 LPUART7_RX CCM_PMIC_RDY CSI_DATA21 GPIO3_IO18 ENET2_TX_EN
So in normal GPIO mode this is GPIO3 pin 18.
But assuming you are using the standard core all of the GPIO pins are switched to be in high speed mode: So this one is on GPIO08 pin 18
And all of these high speed GPIO ports all map to one IRQ: IRQ_GPIO6789 = 157, // RT1060 only
The pins are mapped to high speed in startup.c with the lines:
Code:
IOMUXC_GPR_GPR26 = 0xFFFFFFFF;
IOMUXC_GPR_GPR27 = 0xFFFFFFFF;
IOMUXC_GPR_GPR26 = 0xFFFFFFFF;
IOMUXC_GPR_GPR27 = 0xFFFFFFFF;
IOMUXC_GPR_GPR28 = 0xFFFFFFFF;
IOMUXC_GPR_GPR29 = 0xFFFFFFFF;
= 0xFFFFFFFF;
IOMUXC_GPR_GPR29 = 0xFFFFFFFF;
Now you can switch some of the pins back to normal speed by updating the appropriate IOMUXC register here: probably in this case IOMUXC_GPR_GPR28
Probably the bit: 1<<18
If it were running in this mode: Each of the ports have two IRQs, one for the lower 16 bits and one for the higher 16 bits. In this case it would probably be.
IRQ_GPIO3_16_31 = 85,
Which you would need to attachInterruptVector to this one, and then enable the interrupt and set the priority. And you interface method would probably need to
update the appropriate register to clear the status and maybe the cache of it...