请问（有关sl811hs的中断问题）

下面的这段程序应该如何理解？
原来做程序的时候，中断来了只有一个函数响应呀
这里怎么这么多响应函数？


多些多谢
//-----------------------------------------------------------------------------
// USB Interrupt Handlers
// The following functions are called by the USB interrupt jump table.
//-----------------------------------------------------------------------------

// Setup Data Available Interrupt Handler
void ISR_Sudav(void) interrupt 0
{
GotSUD = TRUE; // Set flag
EZUSB_IRQ_CLEAR();
USBIRQ = bmSUDAV; // Clear SUDAV IRQ
}

// Setup Token Interrupt Handler
void ISR_Sutok(void) interrupt 0
{
EZUSB_IRQ_CLEAR();
USBIRQ = bmSUTOK; // Clear SUTOK IRQ
}

void ISR_Sof(void) interrupt 0
{
EZUSB_IRQ_CLEAR();
USBIRQ = bmSOF; // Clear SOF IRQ
}

void ISR_Ures(void) interrupt 0
{
EZUSB_IRQ_CLEAR();
USBIRQ = bmURES; // Clear URES IRQ
}

void ISR_IBN(void) interrupt 0
{
}

void ISR_Susp(void) interrupt 0
{
Sleep = TRUE;
EZUSB_IRQ_CLEAR();
USBIRQ = bmSUSP;
}

void ISR_Ep0in(void) interrupt 0
{
}

void ISR_Ep0out(void) interrupt 0
{
}

void ISR_Ep1in(void) interrupt 0
{
}

void ISR_Ep1out(void) interrupt 0
{
int i;

for(i=0;i<OUT1BC;i++)
HOSTCMD = OUT1BUF;

OUT1BC = 0;
BULK_OUT_DONE = TRUE;
  EZUSB_IRQ_CLEAR();
   OUT07IRQ = bmEP1;
}

void ISR_Ep2in(void) interrupt 0
{
}

void ISR_Ep2out(void) interrupt 0
{
}

void ISR_Ep3in(void) interrupt 0
{
}

void ISR_Ep3out(void) interrupt 0
{
}

void ISR_Ep4in(void) interrupt 0
{
}

void ISR_Ep4out(void) interrupt 0
{
}

void ISR_Ep5in(void) interrupt 0
{
}

void ISR_Ep5out(void) interrupt 0
{
}

void ISR_Ep6in(void) interrupt 0
{
}

void ISR_Ep6out(void) interrupt 0
{
}

void ISR_Ep7in(void) interrupt 0
{
}

void ISR_Ep7out(void) interrupt 0
{
}

我也有同感，是不是有这么多的中断可以产生呢？