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单片机课程设计 题目1：8*8LED扫描输出 要求： 准备知识： 硬件：J46为列控制，连P1, LD-1为左边开始，J47为行控制，LD-A开始为自上而下，JP2为电源控制。 有关实验： 串并转换实验 参考程序：显示特定图形 汇编： CLK EQU P3.2 DINA EQU P3.3 DINB EQU P3.4 CLEAR EQU P3.5 org 8000h ajmp main org 8100h main: mov sp,#70h MOV 50H,#07EH MOV 51H,#0BDH MOV 52H,#0DBH MOV 53H,#0E7H MOV 54H,#0E7H MOV 55H,#0DBH MOV 56H,#0BDH MOV 57H,#07EH CLR CLEAR SETB CLK SETB DINA SETB DINB SETB CLEAR mainloop: mov r6,#8h mov dptr,#09000h MOV R1,#50H flashcy: MOV A,@R1 MOV R0,A INC R1 clr a movc a,@a+dptr mov P1,0FFH lcall sendto mov P1,a INC DPTR LCALL DELAY DJNZ R6,flashcy sjmp mainloop sendto: PUSH ACC CLR CLK SENDTIME: MOV R7,#08H MOV A,R0 clr c SENDCY: RRC A MOV DINA,C SETB CLK CLR CLK DJNZ R7,SENDCY POP ACC RET DELAY: MOV R7,#01 DELAY0: MOV R4,#17 DELAY1: MOV R3,#28 DJNZ R3,$ DJNZ R4,DELAY1 DJNZ R7,DELAY0 RET org 09000h db 0FEh,0FDh,0FBh,0F7h,0EFh,0DFh,0BFh,07Fh end /*C: */ #include reg52.h sbit CLK =P3^2; sbit DINA =P3^3; sbit DINB =P3^4; sbit CLEAR =P3^5; unsigned char code dispdata[8]={0x7e,0xBD,0xDB,0xE7,0xE7,0xDB,0xBD,0x7E}; unsigned char code dispbit[8]={0xfe,0xfd,0xfb,0xf7,0xef,0xDf,0xBf,0x7f}; bdata unsigned char kdat; sbit cc=kdat^0; void sendto(unsigned char dat); void DELAY(); main() { unsigned char i; for(i=0;i200;i++); CLEAR=0; CLK=1; DINA=1; DINB=1; CLEAR=1; while(1) { for(i=0;i8;i++) { P1=0XFF; sendto(dispdata[i]); P1=dispbit[i]; DELAY(); } } } void sendto(unsigned char dat) { unsigned char i; CLK=0; kdat=dat; for(i=0;i8;i++) { DINA=cc; CLK=1; CLK=0; kdat=kdat1; } } void DELAY() { unsigned char k,j; for(k=0;k10;k++) for(j=0;j50;j++); }