I have a doubt about the C-code of the dual motor demo board for MPC5643L,about the Address of FOC_Fast_ISR

[edenli]

Hidumitru-daniel.popa‌,

Recently,i am learning the dual motor demo board C-code for MPC5643L,I am combing the control process,now i have a question about it:

I right click the mouse on the function 'void FOC_Fast_ISR(void)' .I jump into  a segment c-code in the file named INTCISRVEC_table.c:

#ifdef IRQ_206_FCN /* ISR 206 Equal to FOC_Fast_ISR() */
(tU32)&IRQ_206_FCN,
#else
(tU32)&dummy,
#endif

then Right the 'IRQ_206_FCN' in the INTCISRVEC_table.c

jump into  a segment c-code in the file named MPC5643L_appconfig.h:

#define IRQ_206_FCN                    FOC_Fast_ISR

But i Right The  FOC_Fast_ISR in the  MPC5643L_appconfig.h,it can not jump,

I have a question that ,Where can i find the address of FOC_Fast_ISR  ?

it shouldn't like as  #define  FOC_Fast_ISR   0x12xx xxxx xxxx ?

[Daniel_Popa]

You will find it as an entry in the interrupt table. 

const uint32_t __attribute__ ((section (".intc_vector_table"))) IntcIsrVectorTable[] = { 

...

#ifdef IRQ_206_FCN /* ISR 206 */
(tU32)&IRQ_206_FCN,
#else
(tU32)&dummy,
#endif

}

According with MPC5643L the IRQ_206 is triggered by the ADC Interrupt.

Hope this helps!

Daniel