ECC double bits error interrupt routine test

814420552 — Fri, 07 Aug 2020 02:53:12 GMT

hello .

for FTFC ECC doule bits error interrupt toutine test code is as the following

FTFC->FERCNFG |= 0x02;//DFDIE
FTFC->FERCNFG |= 0x20; //FDFD
INT_SYS_EnableIRQ(FTFC_Fault_IRQn);

i = *(uint32_t *)0x6000;

and interrupt entry function is

void FTFC_Fault_IRQHandler(void)
{
FTFC->FERSTAT |= 0x20;

}

when Flash boot after execute INT_SYS_EnableIRQ(FTFC_Fault_IRQn); it will enter FTFC_Fault_IRQHandler. but if RAM boot ,execute the same code it will not enter FTFC_Fault_IRQHandler.

why is this different between two boot mode?

comment: my device is S32K146 and my compile tool is S32K DS.

Re: ECC double bits error interrupt routine test

danielmartynek — Mon, 10 Aug 2020 12:36:38 GMT

Hello 814420552@qq.com

When FDFD = 1, any valid flash read access sets the DFDIF flag.
I guess the INT_SYS_EnableIRQ() function actually sets the flag when the function is in the flash, not the pointer.
Please read the DFDIF flag in EmbSys_Register view while stepping through the code.

You can put the function before this line:
FTFC->FERCNFG |= 0x20; //FDFD

Also, the i variable might be optimized out by the compiler.

Regards,

Daniel

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ECC double bits error interrupt routine test

Re: ECC double bits error interrupt routine test