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fiq interrupt

hi,

i am using lpc2119 arm controller. i have tried to write a program for fiq interrupt using eint1 external interrup.
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
//program for fiq interrupt
#include <LPC21xx.H> // LPC21xx definitions
void fiqint(void);
int main(void)
{

IODIR1 =0X00FF0000;
PINSEL0 =0X20000000; //SELECT EINT1 IN PINCONNECT BLOCK
VICIntSelect=0x00008000; //ENABLE VIC CHANNEL AS FIQ
VICIntEnable=0x00008000; //ENABLE EINT1 INTERRUPT IN THE VIC
IOCLR1 =0X00FF0000;
while(1);
}

void fiqint(void) __fiq
{

IOSET1 =0X00FF0000;

}
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

above is the program that i have written. i am using keil compiler. when i build the program it is showing an error.

Build target 'Target 1'
assembling Startup.s...
compiling fiq intr.c...
fiq intr.c(13): error: #130: expected a "{"
Target not created

line 13 :void fiqint(void) __fiq

anybody please help me in clarrifying this problem. i am referring 'the insiders guide to the philips arm7 based controller'.hope you will be able correct the error in the code

thanking you
sanoop

Parents
  • You should already have the required manual locally on your machine.

    C:\Keil\ARM\Hlp\RV_cc.pdf

    The phrase that you need to search for within this PDF is Function qualifiers

    This tells you clearly what the correct qualifier is. (It doesn't appear that this qualifier takes advantage of the extra banked registers in fiq mode though)

Reply
  • You should already have the required manual locally on your machine.

    C:\Keil\ARM\Hlp\RV_cc.pdf

    The phrase that you need to search for within this PDF is Function qualifiers

    This tells you clearly what the correct qualifier is. (It doesn't appear that this qualifier takes advantage of the extra banked registers in fiq mode though)

Children
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