如果一個程序中，有兩個中斷，假若為CAP1中斷和CAP2中斷，我查到DSP默認的優先級是CAP2級別高于CAP1，
是不是當CAP2滿足中斷條件后，CAP2進入中斷，即使此刻CAP1也滿足中斷條件，也不會進入CAP1中斷？？
還有就是如果自始至終，CAP1和CAP2都滿足中斷條件的話，CAP1就會一直進不去中斷？？
我在實際測量中發現，分別用兩個信號發生器 都產生兩個50hz的 高低電平信號，幅值滿足DSP能夠進入中斷的要求。
測試中發現，當僅有CAP1中斷時，可以正常捕獲數值，CAP1的兩個寄存器差值為50HZ對應數字量，當僅有CAP2中斷時，也可以正常捕獲數值，CAP2的兩個寄存器差值與CAP1基本差別不大。由此可以確定兩個CAP中斷沒有問題。
但是，當CAP1和CAP2同時滿足中斷條件時，根據程序單步運行發現，兩個中斷也都能進入，但是捕獲數值不正確。此時CAP1和CAP2的兩個寄存器雖然數值時刻在變化，但是差值始終為0，一直弄不清原因，求救中...............
主要程序如下：
#include "DSP280x_Device.h" // DSP280x Headerfile Include File
#include "DSP280x_Examples.h" // DSP280x Examples Include File
interrupt void ecap1_isr(void);
interrupt void ecap2_isr(void);
void InitECapture(void);
void Fail(void);
// Global variables used in this example
Uint32 ECap1IntCount;
Uint32 ECap2IntCount;
Uint32 TSt1;
Uint32 TSt2;
Uint32 TSt3;
Uint32 TSt4;
Uint32 Period1;
Uint32 Period2;
void main(void)
{
InitSysCtrl();
InitECap1Gpio();
InitECap2Gpio();
DINT;
InitPieCtrl();
IER = 0x0000;
IFR = 0x0000;
InitPieVectTable();
EALLOW;
PieVectTable.ECAP1_INT = &ecap1_isr;
PieVectTable.ECAP2_INT = &ecap2_isr;
EDIS;
InitECapture();
ECap1IntCount = 0;
ECap2IntCount = 0;
IER |= M_INT4;
PieCtrlRegs.PIEIER4.bit.INTx1 = 1;
PieCtrlRegs.PIEIER4.bit.INTx2 = 1;
EINT;
ERTM;
for(;;)
{
asm(" NOP");
}
}
void InitECapture()
{
ECap1Regs.ECEINT.all = 0x0000;
ECap1Regs.ECCLR.all = 0xFFFF;
ECap1Regs.ECCTL1.bit.CAPLDEN = 0;
ECap1Regs.ECCTL2.bit.TSCTRSTOP = 0;
ECap1Regs.ECCTL2.bit.CONT_ONESHT = 1;
ECap1Regs.ECCTL2.bit.STOP_WRAP = 3;
ECap1Regs.ECCTL1.bit.CAP1POL = 1;
ECap1Regs.ECCTL1.bit.CAP2POL = 1;
ECap1Regs.ECCTL1.bit.CAP3POL = 1;
ECap1Regs.ECCTL1.bit.CAP4POL = 1;
ECap1Regs.ECCTL1.bit.CTRRST1 = 0;
ECap1Regs.ECCTL1.bit.CTRRST2 = 0;
ECap1Regs.ECCTL1.bit.CTRRST3 = 0;
ECap1Regs.ECCTL1.bit.CTRRST4 = 0;
ECap1Regs.ECCTL2.bit.SYNCI_EN = 0;
ECap1Regs.ECCTL2.bit.SYNCO_SEL = 2;
ECap1Regs.ECCTL1.bit.CAPLDEN = 1;
ECap1Regs.ECCTL2.bit.TSCTRSTOP = 1;
ECap1Regs.ECCTL2.bit.REARM = 1;
ECap1Regs.ECCTL1.bit.CAPLDEN = 1;
ECap1Regs.ECEINT.bit.CEVT4 = 1;
ECap2Regs.ECEINT.all = 0x0000;
ECap2Regs.ECCLR.all = 0xFFFF;
ECap2Regs.ECCTL1.bit.CAPLDEN = 0;
ECap2Regs.ECCTL2.bit.TSCTRSTOP = 0;
ECap2Regs.ECCTL2.bit.CONT_ONESHT = 1;
ECap2Regs.ECCTL2.bit.STOP_WRAP = 3;
ECap2Regs.ECCTL1.bit.CAP1POL = 1;
ECap2Regs.ECCTL1.bit.CAP2POL = 1;
ECap2Regs.ECCTL1.bit.CAP3POL = 1;
ECap2Regs.ECCTL1.bit.CAP4POL = 1;
ECap2Regs.ECCTL1.bit.CTRRST1 = 0;
ECap2Regs.ECCTL1.bit.CTRRST2 = 0;
ECap2Regs.ECCTL1.bit.CTRRST3 = 0;
ECap2Regs.ECCTL1.bit.CTRRST4 = 0;
ECap2Regs.ECCTL2.bit.SYNCI_EN = 0;
ECap2Regs.ECCTL2.bit.SYNCO_SEL = 2;
ECap2Regs.ECCTL1.bit.CAPLDEN = 1;
ECap2Regs.ECCTL2.bit.TSCTRSTOP = 1;
ECap2Regs.ECCTL2.bit.REARM = 1;
ECap2Regs.ECCTL1.bit.CAPLDEN = 1;
ECap2Regs.ECEINT.bit.CEVT4 = 1;
}
interrupt void ecap1_isr(void)
{
TSt1 = ECap1Regs.CAP1;
TSt2 = ECap1Regs.CAP2;
Period1 = TSt2-TSt1;
ECap1IntCount++;
ECap1Regs.ECCLR.bit.CEVT4= 1;
ECap1Regs.ECCLR.bit.INT = 1;
ECap1Regs.ECCTL2.bit.REARM = 1;
PieCtrlRegs.PIEACK.all = PIEACK_GROUP4;
}
interrupt void ecap2_isr(void)
{
TSt3 = ECap2Regs.CAP1;
TSt4 = ECap2Regs.CAP2;
Period2 = TSt4-TSt3;
ECap2IntCount++;
Period4=abs(TSt3-TSt1);
ECap2Regs.ECCLR.bit.CEVT4 = 1;
ECap2Regs.ECCLR.bit.INT = 1;
ECap2Regs.ECCTL2.bit.REARM = 1;
PieCtrlRegs.PIEACK.all = PIEACK_GROUP4;
}
void Fail()
{
asm(" ESTOP0");
}
//===========================================================================
// No more.
//===========================================================================
【問】2806 CAP和中斷
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我只能頂起這個問題，請版主去回答！貢獻自己一點力量啦