66AK2L06 FFTC_0和FFTC_1 与DSP 之间的 QUEUE channel 是多少？

各位老师好：

        我在将例程 fftcMCK2LC66BiosExampleProject 从一个FFTC处理数据 修改为 FFTC_0和FFTC_1同时处理数据时，RX打不开，可能是RXOBJ_1配置中rxCfg_1.cppiRxQNum= myQNum; 错误，所以 FFTC_1 与DSP 之间的 QUEUE channel 是多少 ？ 希望老师给够详细解答。

rxobj_1配置代码如下：

switch (coreNum)

{

case 0: {
       myQNum = 712;
       destQNum = 713;
       accChanNum = FFTC_RX_ACC_CHANNEL_NUM;        break;
}
case 1: {
       myQNum = 713;
       destQNum = 714;
       accChanNum = FFTC_RX_ACC_CHANNEL_NUM + 1;
       break;
}
case 2: {
       myQNum = 714;
       destQNum = 715;
       accChanNum = FFTC_RX_ACC_CHANNEL_NUM + 2;
       break;
}
case 3: {
       myQNum = 715;
       destQNum = 712;
       accChanNum = FFTC_RX_ACC_CHANNEL_NUM + 3;
       break;
}
default:
goto error;

}

memset (&rxCfg_1, 0, sizeof (rxCfg_1));
rxCfg_1.useFlowId = -1;
rxCfg_1.bManageRxFlowCfg = 1; /* Let the driver do Rx FDQ management */
rxCfg_1.rxFlowCfg.drvCfg.descType = Cppi_DescType_HOST;
rxCfg_1.rxFlowCfg.drvCfg.cppiNumDesc = 2;
rxCfg_1.rxFlowCfg.drvCfg.bufferSize = (pFFTAppCfg->numOutputSamples * FFTC_SAMPLE_SIZE + 4) * pFFTAppCfg->numBlocks + 4;
rxCfg_1.rxFlowCfg.drvCfg.bPSInfoPresent = 0;
rxCfg_1.rxFlowCfg.drvCfg.psLocation = Cppi_PSLoc_PS_IN_SOP;
rxCfg_1.cppiRxQNum = myQNum;
rxCfg_1.bUseInterrupts = 1;
rxCfg_1.bManageAccumList = 1; /* Let driver do the ISR/Accum handling */
rxCfg_1.accumCfg.drvCfg.bEnablePacing = 0;
rxCfg_1.accumCfg.drvCfg.intThreshold = 1;
rxCfg_1.accumCfg.drvCfg.accChannelNum = accChanNum;
rxCfg_1.bBlockOnResult = 1;
if ((hRxObj_1 = Fftc_rxOpen (hFFTC_0, &rxCfg_1)) == NULL)
{
       System_printf ("[Core %d]: 1 Rx object open failed \n", coreNum);
       goto error;
}
else
{
       System_printf ("[Core %d]: 1 Rx flow %d opened successfully using Rx queue %d \n", coreNum,
       Fftc_rxGetFlowId (hRxObj_1), Fftc_rxGetRxQueueNumber (hRxObj_1));
}

Shine：

请参考e2echina.ti.com/…/477931

kang xiao：

回复 Shine:

谢谢你的回答

1.为什么程序中 对于case 0: myQNum = 712; destQNum = 713; accChanNum = FFTC_RX_ACC_CHANNEL_NUM(16);  ?

2.既然TX QUEUE  Queues 688 to 691 are attached to FFTC‐1，为什么例程中  Tx queue = 0 ; ?

/* Lets run 16 sample size request using FFTC Tx queue 0. */  queueNum = 0;

/* Run the example */  fftc_run_example (coreNum, queueNum);