Hi I am Hiroki.
I have any open issue with IRQ0 - I am working with ADSP-21488.I want to configure this pin as hardware interrupt and the process I did is as follows:
*pSYSCTL |= IRQ0EN;asm（ "bit set mode2 IRQ0E;"）; // EDGE SENSITIVEinterrupt( SIG_IRQ0, MyFunc );sysreg_bit_set( sysreg_MODE1, IRPTEN );
Use the debugger inside main-When an interrupt occurs, the program jumps to the beginning of main instead of MyFunc outside main.Note that even if a breakpoint is set in MyFunc during debugging, it does not move to MyFunc .If the IRPTL register is monitored during debugging, “IRQ0I”, which is the 10th bit of the IRPTL register, becomes “1” at the same time that IRQ0 is received. After that, when jumps to the beginning of main, “IRQ0I” becomes “0”.what is my mistake?
Hi Hiroki,From your mail, I understand that you wish to use hardware interrupt but facing issue with this. Please correct if my understanding is wrong.I've a push button interrupt code in my repository. This can be directly run on ADSP-21489 EZ-kit. Whenever push-button 1 is pressed, interrupt occurs and corresponding ISR is executed. Please use this code as a reference and modify as per your requirement.Hope this helps!Regards,Anand Selvaraj.DPI Interrupt.zip
Hi Anand Selvaraj
IRQ0 interrupt was successful with reference to the attached file.Thank you very much.
SHARC Processor Programming Reference (Revision 2.4) Figure 4-4. Interrupt Process Flow is also helpful.