51 單片機編程：兩路兩位的計數(shù)器
利用?51?單片機構(gòu)成計數(shù)器，計數(shù)的上限，做的大一些，也不難，幾千幾萬都可以。
下面的電路，小了一點，是個兩路兩位的計數(shù)器。
用來記錄乒乓球比賽，還算可以，用于籃球比賽，分?jǐn)?shù)超過?99，這個電路就不夠用了。
電路圖中，用的是共陽數(shù)碼管；還設(shè)置了六個按鍵，用途，都已經(jīng)標(biāo)出。
用?C?語言編寫驅(qū)動程序，是比較簡單的，全部代碼如下：
#include
unsigned?char??n1,?n2;
//-----------------------------------------------
void?delay(unsigned?int?i)???//1ms延時程序
{
????unsigned?char??j;
????while(i--)???for(j?=?0;?j?<?115;?j++);
}
//-----------------------------------------------
void?disp(void)
{
????unsigned?char?code?SEG[]?=?{
??????0xc0,?0xf9,?0xa4,?0xb0,?0x99,?//0-4
??????0x92,?0x82,?0xf8,?0x80,?0x90};//5-9
????P0?=?SEG[n1?/?10];?P1?=?1;?delay(2);?P1?=?0;
????P0?=?SEG[n1?%?10];?P1?=?2;?delay(2);?P1?=?0;
????P0?=?SEG[n2?/?10];?P1?=?4;?delay(2);?P1?=?0;
????P0?=?SEG[n2?%?10];?P1?=?8;?delay(2);?P1?=?0;
}
//-----------------------------------------------
void?main(void)
{
????unsigned?char?k,?t;
????n1?=?n2?=?0;
????while(1)?{
??????disp();
??????if?(~(P2?|?0xc0))?{
????????disp();
????????if?(~(P2?|?0xc0))?{
??????????k?=?~(P2?|?0xc0);
??????????while?(~(P2?|?0xc0))??disp();
??????????if???????(k?==?1)??{n1++;?if?(n1?==?100)?n1?=?0;}
??????????else??if?(k?==?2)??{n1--;?if?(n1?==?-1)?n1?=?99;}
??????????else??if?(k?==?4)??{n2++;?if?(n2?==?100)?n2?=?0;}
??????????else??if?(k?==?8)??{n2--;?if?(n1?==?-1)?n2?=?99;}
??????????else??if?(k?==?16)?{n1?=?n2?=?0;}
??????????else??if?(k?==?32)?{t?=?n1;?n1?=?n2;?n2?=?t;}
????}?}?}
}
//-----------------------------------------------
是不是很簡單？
用匯編語言來編程，就顯得長多了。
和上述?C?語言功能相同的匯編語言程序，全部代碼如下：
????ORG???0000H
????JMP???MAIN
????ORG???0030H
;-------------------------
MAIN:
????CALL??DISP
????MOV???A,?P2
????CJNE??A,?#0FFH,?KEY
????CLR???00H
????JMP???MAIN
;-------------------------
KEY:
????JNB???00H,?A_ADD1
????JMP???MAIN
;----------------------
A_ADD1:
????SETB??00H
????JB????P2.0,?A_SUB1
????MOV???A,?R0
????ADD???A,?#01H
????DA????A
????MOV???R0,?A
????JMP???MAIN
;----------------------
A_SUB1:
????JB????P2.1,?B_ADD1
????MOV???A,?R0
????ADD???A,?#99H
????DA????A??????
????MOV???R0,?A
????JMP???MAIN
;----------------------
B_ADD1:
????JB????P2.2,?B_SUB1
????MOV???A,?R1
????ADD???A,?#01H
????DA????A
????MOV???R1,?A
????JMP???MAIN
;----------------------
B_SUB1:
????JB????P2.3,?CLR_0
????MOV???A,?R1
????ADD???A,?#99H
????DA????A
????MOV???R1,?A
????JMP???MAIN
;----------------------
CLR_0:
????JB????P2.4,?EXCH
????MOV???R0,?#00H
????MOV???R1,?#00H
????JMP???MAIN
;----------------------
EXCH:
????JB????P2.5,?END_K
????MOV???A,?R0
????XCH???A,?R1
????MOV???R0,?A
????JMP???MAIN
END_K:
????CLR???00H
????JMP???MAIN
;-------------------------
DISP:
????MOV???DPTR,?#BG
????MOV???A,?R0
????SWAP??A
????ANL???A,?#0FH
????MOVC??A,?@A?+?DPTR
????MOV???P0,?A
????SETB??P1.0
????CALL??DELAY
????CLR???P1.0
;----------------------
????MOV???A,?R0
????ANL???A,?#0FH
????MOVC??A,?@A?+?DPTR
????MOV???P0,?A
????SETB??P1.1
????CALL??DELAY
????CLR???P1.1
;----------------------
????MOV???A,?R1
????SWAP??A
????ANL???A,?#0FH
????MOVC??A,?@A?+?DPTR
????MOV???P0,?A
????SETB??P1.2
????CALL??DELAY
????CLR???P1.2
;----------------------
????MOV???A,?R1
????ANL???A,?#0FH
????MOVC??A,?@A?+?DPTR
????MOV???P0,?A
????SETB??P1.3
????CALL??DELAY
????CLR???P1.3
RET
;-------------------------
DELAY:
????DJNZ??R7,?$
????DJNZ??R7,?$
????DJNZ??R7,?$
????DJNZ??R7,?$
RET
;-------------------------
BG:
????DB??0C0H,0F9H,0A4H,0B0H,?99H
????DB???92H,?82H,0F8H,?80H,?90H
;-------------------------
END
匯編語言的程序，看起來，比?C?語言的程序，長了不少，但是，編譯后生成的機器碼，卻少于?C?程序。
誰來看一看，編譯后生成的機器碼占用?ROM?空間，相差了多少？
其實，這個匯編語言的程序，編程的思路，并不是很好的，程序，還是可以優(yōu)化的。
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