[SOLVED] lpc2148 uart problem

 #include "LPC214x.H"    // LPC2148 MPU Register
#include <stdio.h>                                 // For Used Function printf

/* pototype  section */
void init_serial0 (void);                            // Initil UART-0
                                    // ADC Result Volt
int main(void)
{ 
  init_serial0();                                    // Initial UART0 = 9600,N,8,1
  printf("Input Voltage 0-3.3V to P0.6 For Test\n");       // Call prinff Function
  // xxxx xxxx xxxx xxxx xx11 xxxx xxxx xxxx
}
void init_serial0 (void) 
{
  PINSEL0 &= 0xFFFFFFF0;                           // Reset P0.0,P0.1 Pin Config
  PINSEL0 |= 0x00000001;                           // Select P0.0 = TxD(UART0)
  PINSEL0 |= 0x00000004;                           // Select P0.1 = RxD(UART0)

  U0LCR &= 0xFC;                                 // Reset Word Select(1:0)
  U0LCR |= 0x03;                                 // Data Bit = 8 Bit
  U0LCR &= 0xFB;                                 // Stop Bit = 1 Bit
  U0LCR &= 0xF7;                                 // Parity = Disable
  U0LCR &= 0xBF;                                 // Disable Break Control
  U0LCR |= 0x80;                                 // Enable Programming of Divisor Latches

  // U0DLM:U0DLL = 60.00 MHz / [16 x Baud]
  //             = 60.00 MHz / [16 x 9600]
  //             = 390.6 = 391 = 0187H
  U0DLM = 0x01;                                    // Program Divisor Latch(391) for 9600 Baud
  U0DLL = 0x87;

  U0LCR &= 0x7F;                                 // Disable Programming of Divisor Latches

  U0FCR |= 0x01;                                 // FIF0 Enable
  U0FCR |= 0x02;                                 // RX FIFO Reset
  U0FCR |= 0x04;                                 // TX FIFO Reset
  U0FCR &= 0x3F;                     
}