Hello experts,
I am using SAM L11 (Core is Cortex-M23).I did a simple test of the interrupt.In the case of the handler was the secure world, it worked as I expected.The instruction sequences are as the follows. *(int*)(0xE000E380) = 0; *(int*)(0xE000E100) = 1; *(int*)(0xE000E200) = 1; for(i=0;i<1000;i++);However, when Interrupt Target Non-secure Register (0xE000E380) was set,the HardFault had occurred. As for SAM L11, would it be reasonable behavior? *(int*)(0xE000E380) = 1; *(int*)(0xE000E100) = 1; *(int*)(0xE000E200) = 1; /* HardFault */ for(i=0;i<1000;i++);Although I put the vector of the secure world to the non-secure world address, the phenomenon was not changed.From the specification aspect, how does it work?
Thank you and best regards,
Yasuhiko Koumoto.
Hi there,
I think if you target the interrupt to the non-secure world, and you should enable the interrupt in the non-secure world but not the secure world. If you open it in secure world, there will be a fault. So you can have a try.
Best regards.