Question

问题：我使用ARM汇编语言对LPC2104微处理器进行编程，我只是试图制作一个接收UART输入并将其传回的程序。它只是从内存中传输一个字符串就可以了，但我通过添加一个接收子程序并将内存中的字符串带走来修改它。问题必须在于接收子程序，因为我甚至无法输入UART1。我还没有编辑＆＃34; loop＆＃34;的内容。正确的，所以现在就忽视这一点。这是我在ARM组装或微处理器上工作的第一个学期，所以请向外行解释。

这是我的代码：

        AREA uartdemo, CODE, READONLY
pinsel0    EQU  0xE002C000; controls function of pins
u0start    EQU  0xE000C000; start of UART0 registers
lcr0       EQU  0xC;        line control register for UART0
lsr0       EQU  0x14;       line status register for UART0
ramstart   EQU  0x40000000; start of onboard RAM for 2104
stackstart EQU  0x40000200; start of stack

ENTRY

start
    LDR sp, =stackstart; set up stack pointer
    BL  UARTConfig;      initialize/configure UART0
loop
    BLNE Receive;        receive character from UART
    BLNE Transmit;       send character to UART
    BNE  loop;           continue if not a '0'
done B done;             otherwise, we are done             

; subroutine UARTConfig
;   Configures the I/O pins first.
;   Then sets up the UART control register.
;   Parameters set to 8 bits, no parity, and 1 stop bit.
;   Registers used:
;   r5 - scratch register
;   r6 - scratch register
;   inputs:  none
;   outputs:  none
UARTConfig
    PUSH {r5, r6, lr};
    LDR  r5, =pinsel0;    base address of register
    LDR  r6, [r5];        get contents
    BIC  r6, r6, #0xf;    clear out lower nibble
    ORR  r6, r6, #0x5;    sets P0.0 to Tx0 and P0.1 to Rx0
    STR  r6, [r5];        r/modify/w back to register
    LDR  r5, =u0start;
    MOV  r6, #0x83;       set 8 bits, no parity, 1 stop bit
    STRB r6, [r5, #lcr0]; write control byte to LCR
    MOV  r6, #0x61;       9600 baud @ 15MHz VPB clock
    STRB r6, [r5];        store control byte
    MOV  r6, #3;          set DLAB = 0
    STRB r6, [r5, #lcr0]; Tx and Rx buffers set up
    POP  {r5, r6, pc};

; subroutine Transmit
;   Puts one byte into the UART for transmitting
;   Registers used:
;   r5 - scratch register
;   r6 - scratch register
;   inputs:  r0 - byte to transmit
;   outputs:  none
Transmit
     PUSH {r5, r6, lr};
     LDR  r5, =u0start;
wait LDRB r6, [r5, #lsr0];  get status of buffer
     TST  r6, #0x20;        buffer empty?
     BEQ wait;              spin until buffer is empty
     STRB r0, [r5];
     POP {r5, r6, pc};

; subroutine Receive
;   Takes one byte from the UART for transmitting
;   Registers used:
;   r5 - scratch register
;   r6 - scratch register
;   inputs:  r0 - byte to transmit X
;   outputs:  none                 X
Receive
     PUSH {r5, r6, lr};
     LDR  r5, =u0start;
wait2 LDRB r6, [r5, #lsr0];  get status of buffer
     TST  r6, #0x00;        buffer empty?
     BEQ wait2;              spin until buffer is empty
     STRB r0, [r5];
     POP {r5, r6, pc};

    END

来自Keil的UART与2104相呼应

0 个答案: