Hi.
The following structure is declared in imx-scu.c (/ kernel_imx / drivers / firmware / imx).
struct imx_sc_ipc {
/ * SCU uses 4 Tx and 4 Rx channels * /
struct imx_sc_chan chans [SCU_MU_CHAN_NUM];
struct device * dev;
struct mutex lock;
struct completed done;
bool fast_ipc;
/ * temporarily store the SCU msg * /
u32 * msg;
u8 rx_size;
u8 count;
};
It says "SCU uses 4 Tx and 4 Rx channels", but I think there are 0 to 4 channels.
Isn't it a total of 10 channels (5 Tx and 5 Rx channels)?
"SCU_MU_CHAN_NUM" is "8" by #define.
What should I fix if I want to use all 5 channels?
Please let me know if I am wrong.
Solved! Go to Solution.
Hi @sushi_happy
It says "SCU uses 4 Tx and 4 Rx channels", but I think there are 0 to 4 channels. Isn't it a total of 10 channels (5 Tx and 5 Rx channels)?
> The MU has four, 32-bit write-only transmit registers and four, 32-bit read-only receive registers on the Processor B and Processor A-sides. They are not like 0 to 4 channels. Actually with 4 registers for each side.
What should I fix if I want to use all 5 channels?
> 4 Channels by default are designed according to the IP hardware and SoC.
If you want to know more details, I would recommend you to check the 19.9 Messaging Unit (MU) of imx8qm reference manual.
Best regards
Harvey
Hi @sushi_happy
It says "SCU uses 4 Tx and 4 Rx channels", but I think there are 0 to 4 channels. Isn't it a total of 10 channels (5 Tx and 5 Rx channels)?
> The MU has four, 32-bit write-only transmit registers and four, 32-bit read-only receive registers on the Processor B and Processor A-sides. They are not like 0 to 4 channels. Actually with 4 registers for each side.
What should I fix if I want to use all 5 channels?
> 4 Channels by default are designed according to the IP hardware and SoC.
If you want to know more details, I would recommend you to check the 19.9 Messaging Unit (MU) of imx8qm reference manual.
Best regards
Harvey
Thank you, @Harvey021
I understand that this description refers to the MU of the SCU.
I thought it was a description of LPUART of DMA.