LPC2294 UART Interrupt problem

thanks my friend,

/*INIT*/
void serial_init (void) {
// PINSEL0 = 0x00050000; /* Enable RxD1 and TxD1
*/
PINSEL0 |= 0x00000005; /// UART0
U0FCR=0x01; ///Enable FIFOS and reset
U0LCR = 0x83; ///DLAB=1
U0DLL = 0x20; /// 115200 baud
U0DLM = 0x00;
U0LCR = 0x03; /// 8 bits, no Parity, 1 Stop
bit ve DLAB=0 yapılıyor */
U0IER = 0x03; /// Enable RDA and THRE
interrupts */
VICVectAddr14 = (U32)int_serial; /// 14th vector is serial interrupt
VICVectCntl14 = 0x27; /// enable vect14 and assign IRQ number 7
VICIntEnable |= 0x80; /// IRQ 7 enable
}

thanks my friend,

/*INIT*/
void serial_init (void) {
// PINSEL0 = 0x00050000; /* Enable RxD1 and TxD1
*/
PINSEL0 |= 0x00000005; /// UART0
U0FCR=0x01; ///Enable FIFOS and reset
U0LCR = 0x83; ///DLAB=1
U0DLL = 0x20; /// 115200 baud
U0DLM = 0x00;
U0LCR = 0x03; /// 8 bits, no Parity, 1 Stop
bit ve DLAB=0 yapılıyor */
U0IER = 0x03; /// Enable RDA and THRE
interrupts */
VICVectAddr14 = (U32)int_serial; /// 14th vector is serial interrupt
VICVectCntl14 = 0x27; /// enable vect14 and assign IRQ number 7
VICIntEnable |= 0x80; /// IRQ 7 enable
}

I managed to get into the interrupt vector.
I used IRQ6 instead of IRQ7 for UART0. I don't know why and how .. but it works.

Can anybody explain this situation?

I think your problem is VICIntEnable |= 0x80; /// IRQ 7 enable
This is not for IRQ enable but for UART0 interrupt as IRQ enable and it is bit 6 of VICIntEnable .

Once enabled you may assign it to any IRQ.

Sorry. I didn't understand.

UART0 is vercto number 6 not 7.
so,

VICVectCntl14 = 0x26;
VICIntEnable |= 0x40;

Try this.

Thanks
Dimple