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Change polarity of an EINT inside it irq routine

Question asked by Alessandro Melino on Jul 18, 2019
Latest reply on Jul 22, 2019 by soledad

Hello.

 

As you can read in the title, I am trying to change the polarity of the edge which causes the external interrupt. I am using an LPC2129.

 

Firstly I configure it as rising edge sensitive and the VIC vectors registers:

 

PINSEL1 |= (0<<1) | (1<<0);                  //Pin P0.16 as EINT0
EXTMODE |= (1<<0) |;                          //Edge sensitive
EXTPOLAR |= (1<<0) ;                         //Rising edge

 

 

VICVectCntl1 = 0x20 |14;                     
VICVectAddr1 =(unsigned)EINT0_isr; 
VICIntEnable |= (1<<14); 
EXTINT |= (1<<0);

 

Then, in the irq routine I want to change to falling edge sensitive, as the following code:

 

void EINT0_isr (void) __irq { 

 

EXTINT |= (1<<0);           //Clear flag
...
EXTPOLAR &= ~(1<<0); //Change to falling edge

...}

 

This does not work and the program crash.

 

Then I read this in the manual but I do not understand how to do it:

 

Note: Software should only change a bit in this register when its interrupt is
disabled in the VICIntEnable register, and should write the corresponding 1 to the
EXTINT register before enabling (initializing) or re-enabling the interrupt, to clear
the EXTINT bit that could be set by changing the polarity.

 

So I tried to change the code like this but did not work:

 

VICIntEnClr |= (1<<14);     
EXTINT |= (1<<0);
EXTPOLAR &= ~(1<<0);
VICIntEnable |= (1<<14); 

 

Which is the correct way to change edge polarity inside the irq routine of an EINT?

 

Best regards,

Alessandro

 

P.S.: sorry for my english.

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