STM32L486: SPI-Clock doubled

[Byteslave]

Use SPI with MCU above.  With

spi_transmit(0x55);

and

void spi_transmit(uint8_t tx_data)
{
  SPI1->DR= tx_data;
  while (!(SPI1->SR & SPI_SR_TXE));
}

I have 16 SPI clocks. Init was:

hspi1.Init.DataSize = SPI_DATASIZE_8BIT;

 Data on MOSI showing on the first 8 clocks, MOSI on remaining 8 clocks is zero.

While using HAL (same init):

pdat= 0x55;
HAL_SPI_Transmit(&hspi1, (const uint8_t *)&pdat, 1, 100);

I have, as expected, 8 SPI clocks.

Any Suggestions? Thanks!

[TDK · Accepted solution]

Write 8 bits to DR instead of 16.

*(volatile uint8_t*)&SPI1->DR = tx_data;

[Byteslave]

Wow, that worked! I'm still confused, because "tx_data" is 8bits, init was for 8bit data and the subroutine already worked at an STM32F103C8T6.  But it is like it is, Thanks again!

[TDK]

The access size is dictated by the left hand side of the expression and since SPI->DR is a 16-bit register, that's what you write regardless of what is on the right, which gets converted to the right size.

F103 doesn't have data packing so it works there.