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UNRECOGNIZED PRAGMA

AC53351
Intellectual 370 points

Hello

For a particular application I want to use INT5 in 6472. I am using CCS v4. So I'm writing the ISR like this:

 

#pragma vector=5
__interrupt void my_isr(void)
{
    ...................

    return;
}

But it is giving a warning "unrecognized #pragma". How to solve this problem? Do I need to include any header file to make it run?

Regards,

AC.

  • 0
    TI__Guru* 84110 points

    There is no "vector=" pragma available. Not all pragma's are portable from one compiler to another.

    If you will be using DSP/BIOS, which is recommended for simplifying your development, please refer to the DSP/BIOS User's Guide and examples for how to write and configure interrupt handlers.

    If you are not using DSP/BIOS, please look at the CSL examples for how to implement interrupts.