STM32G0 USART3 and USART4 share an interrupt. How to deal with it?

[DRobe.4]

void USART3_IRQHandler(void) is incorrect, since USARTs 3/4 share the interrupt, it doesn't compile. What replaces 'USART3' in the interrupt handler?

NVIC_ClearPendingIRQ(USART3_IRQn); is incorrect since USART3 3/4 share the interrupt. What replaces 'USART3_IRQn' so that the NVIC_ClearPendingIRQ will compile?

[Tesla DeLorean]

You check and service both peripherals and the flags should clear.

I​ think in one of the recent threads it establishes the combined symbol and number for the IRQ. Grep the library source or truncate the Handler name seen in startup.s

[Tesla DeLorean]

YOUR thread here ​https://community.st.com/s/question/0D53W00002FEy7wSAD/i-there-a-usart3irqn-and-a-usart4irqn-on-stm32g070

[KnarfB · Accepted solution]

The interrupt handler names are defined in the startup code, like in startup_stm32g070xx.s:

.word USART1_IRQHandler     /* USART1                 */
.word USART2_IRQHandler     /* USART2                 */
.word USART3_4_IRQHandler  /* USART3, USART4  */

If you create code with UART 3/4 interrupts enabled, you find the code in stm32g0xx_hal_msp.c:

HAL_NVIC_SetPriority(USART3_4_IRQn, 0, 0);
HAL_NVIC_EnableIRQ(USART3_4_IRQn);

Those defines are from CMSIS like stm32g070xx.h.

The generated code is correct and does compile without errors.

hth

KnarfB

[DRobe.4]

Everyone, thanks for your help... compiles, so I presume that it will work.