关于EDMA的一个疑问

lester zhang

如果是手动触发的话，单核能否开启两个channel进行传输？我现在情况是按照例程定义了两个channel，分别映射不同的param，不同的TCC，但是当我第一次手动触发的时候，设置的传递参数是channe1，可是触发完成后发现，中断完成位确是对应的channel2的TCC。后来我发现无论我输入的是channel1 还是channel2，触发的都是channel2 param所对应的传输。。。

12 年多前

0 lester zhang 12 年多前

Intellectual 580 points

if (CSL_edma3Init(&context) != CSL_SOK)
{
printf ("Error: EDMA module initialization failed\n");
}

hModule = CSL_edma3Open(&edmaObj, INSTNUM, NULL, &status);
if ( (hModule == NULL) || (status != CSL_SOK))
{
printf ("Error: EDMA module open failed\n");
}

chAttr.regionNum = CSL_EDMA3_REGION_GLOBAL; /*GLOBAL*/
chAttr.chaNum= 0;
hChannel1 = CSL_edma3ChannelOpen(&chObj, INSTNUM, &chAttr, &status);
if ((hChannel1 == NULL) || (status != CSL_SOK))
{
printf ("Error: Unable to open EDMA Channel:%d\n", 0);
}

chAttr.chaNum= 1;
hChannel2 = CSL_edma3ChannelOpen(&chObj, INSTNUM, &chAttr, &status);
if ((hChannel2 == NULL) || (status != CSL_SOK))
{
printf ("Error: Unable to open EDMA Channel:%d\n", 1);
}

if (CSL_edma3HwChannelSetupQue(hChannel1,CSL_EDMA3_QUE_0) != CSL_SOK)
{
printf ("Error: EDMA channel setup queue failed\n");
}

if (CSL_edma3HwChannelSetupQue(hChannel2,CSL_EDMA3_QUE_0) != CSL_SOK)
{
printf ("Error: EDMA channel setup queue failed\n");
}

CSL_edma3MapDMAChannelToParamBlock (hModule, 0, 0);
CSL_edma3MapDMAChannelToParamBlock (hModule, 1, 1);

Param0= CSL_edma3GetParamHandle(hChannel1, 0, &status);
if (Param0 == NULL)
{
printf ("Error: EDMA Get Parameter Entry failed for paramID.\n");

}

Param1= CSL_edma3GetParamHandle(hChannel2, 1, &status);
if (Param1 == NULL)
{
printf ("Error: EDMA Get Parameter Entry failed for paramID.\n");

}

myParamSetup.option = CSL_EDMA3_OPT_MAKE(CSL_EDMA3_ITCCH_DIS, \
CSL_EDMA3_TCCH_DIS, \
CSL_EDMA3_ITCINT_DIS, \
CSL_EDMA3_TCINT_EN, \
0, CSL_EDMA3_TCC_NORMAL,\
CSL_EDMA3_FIFOWIDTH_NONE, \
CSL_EDMA3_STATIC_DIS, \
CSL_EDMA3_SYNC_AB, \
CSL_EDMA3_ADDRMODE_INCR, \
CSL_EDMA3_ADDRMODE_INCR );

myParamSetup.srcAddr = (Uint32)DDR;//源地址
myParamSetup.aCntbCnt = CSL_EDMA3_CNT_MAKE(4,512);//因为传输的是Uint32，所以每次array跳变是4byte，一共跳变512
myParamSetup.dstAddr = (Uint32)mem;//目的地址，L2地址必须用全局地址
myParamSetup.srcDstBidx = CSL_EDMA3_BIDX_MAKE(4,4);
myParamSetup.linkBcntrld= CSL_EDMA3_LINKBCNTRLD_MAKE(0xFFFF,0);//无link
myParamSetup.srcDstCidx = CSL_EDMA3_CIDX_MAKE(0,1);
myParamSetup.cCnt = 1;

if(CSL_edma3ParamSetup(Param0,&myParamSetup) != CSL_SOK)
{
printf ("Error: EDMA Parameter Entry Setup failed\n");
}

myParamSetup.option = CSL_EDMA3_OPT_MAKE(CSL_EDMA3_ITCCH_DIS, \
CSL_EDMA3_TCCH_DIS, \
CSL_EDMA3_ITCINT_DIS, \
CSL_EDMA3_TCINT_EN, \
1, CSL_EDMA3_TCC_NORMAL,\
CSL_EDMA3_FIFOWIDTH_NONE, \
CSL_EDMA3_STATIC_DIS, \
CSL_EDMA3_SYNC_AB, \
CSL_EDMA3_ADDRMODE_INCR, \
CSL_EDMA3_ADDRMODE_INCR );

myParamSetup.srcAddr = (Uint32)mem;//源地址
myParamSetup.aCntbCnt = CSL_EDMA3_CNT_MAKE(4,512);//因为传输的是Uint32，所以每次array跳变是4byte，一共跳变512
myParamSetup.dstAddr = (Uint32)L2+ (0x10000000);//目的地址，L2地址必须用全局地址
myParamSetup.srcDstBidx = CSL_EDMA3_BIDX_MAKE(4,4);
myParamSetup.linkBcntrld= CSL_EDMA3_LINKBCNTRLD_MAKE(0xFFFF,0);//无link
myParamSetup.srcDstCidx = CSL_EDMA3_CIDX_MAKE(0,1);
myParamSetup.cCnt = 1;

if(CSL_edma3ParamSetup(Param1,&myParamSetup) != CSL_SOK)
{
printf ("Error: EDMA Parameter Entry Setup failed\n");
}

regionIntr.region = CSL_EDMA3_REGION_GLOBAL;
regionIntr.intr = 0x0003; //中断使能位必须和所设置的TCC相同
regionIntr.intrh = 0x0000;

CSL_edma3HwControl(hModule,CSL_EDMA3_CMD_INTR_ENABLE,&regionIntr);

regionIntr.intr = 0x0000;

CSL_edma3HwChannelControl(hChannel1,CSL_EDMA3_CMD_CHANNEL_SET,NULL);

do {
CSL_edma3GetHwStatus(hModule,CSL_EDMA3_QUERY_INTRPEND,&regionIntr);
} while (!(regionIntr.intr & 0x0001));//

CSL_edma3HwControl(hModule,CSL_EDMA3_CMD_INTRPEND_CLEAR,&regionIntr);

CSL_edma3HwChannelControl(hChannel1,CSL_EDMA3_CMD_CHANNEL_SET,NULL);

do {
CSL_edma3GetHwStatus(hModule,CSL_EDMA3_QUERY_INTRPEND,&regionIntr);
} while (!(regionIntr.intr & 0x0002));//

CSL_edma3HwControl(hModule,CSL_EDMA3_CMD_INTRPEND_CLEAR,&regionIntr);

0 lester zhang 12 年多前回复 lester zhang

Intellectual 580 points

2楼是我的代码

0 lester zhang 12 年多前回复 lester zhang

Intellectual 580 points

急啊

0 Andy Yin 12 年多前

TI__Guru 61135 points

注意函数CSL_edma3ChannelOpen的内部实现，其返回值为hEdma = (CSL_Edma3ChannelHandle)pEdmaObj;所以按照你的这种配置顺序，每个channel在调用如下函数open channel时，其输入第一个参数必须不一样，否则返回值只会记录最后一个channel的参数，所以你的代码中在后面enable channel时使能的始终是channel2.

CSL_Edma3ChannelHandle CSL_edma3ChannelOpen
(
CSL_Edma3ChannelObj* pEdmaObj,
CSL_InstNum edmaNum,
CSL_Edma3ChannelAttr* pChAttr,
CSL_Status* pStatus
)

0 lester zhang 12 年多前回复 Andy Yin

Intellectual 580 points

感谢！！

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关于EDMA的一个疑问