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发表于 2018-10-21 20:11:37
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搞明白了怎么显示了,先写个显示的demo显示数字0,为了便于仿真,不停输出,实际使用的时候,是可以锁在的,不改变数字时,不用写入,然后再用proteus仿真一下,因为proteus里没有SAA1060这个元件,要不可直接看数码管显示,现在只能看虚拟仪器里的时序了,代码附后,代码写得效率可能不是很好,但看仿真的结果应该是没有问题。: C( [4 N# H- x' e2 v6 e: I# ~
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# v- B# V! t3 E- Pdemo 的代码在这里# Z3 a( {6 x* T" D7 F% T6 P
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#include "reg52.h" //此文件中定义了单片机的一些特殊功能寄存器- m, a9 G% E' x; a1 n7 B0 ~& K1 W3 a
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sbit yData = P2^0;
3 z. v( O2 H2 M0 N6 rsbit yDLEN = P2^2;
& e& I/ P Q3 @' Usbit yCLK = P2^1;
) l/ X: L- |0 b w9 q, `+ dconst int dailyCount = 200;
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' ?7 R' i2 ?& n1 e. nvoid delay(unsigned int i)/ ~6 A5 x6 Z* U( B; x
{
- Z6 G1 B: G( v while(i--); 2 Y; @/ E0 w# j9 A: [# q& G
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void disp(unsigned int num )) ~3 k6 b0 ^, @0 `5 n! q9 E
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unsigned char i ;3 u6 e3 `9 w0 n
9 K- L2 b, {) g) Z unsigned char dispA[20];! M5 F0 g+ [7 U6 M
unsigned char dispB[20];
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% S: q& G/ l9 P unsigned char num1;6 A* P+ {- d7 b+ N% v7 Q8 v
unsigned char num2;
! c$ @9 P$ n5 a- k( I! E unsigned char num3;) Q" V, ?$ x' l4 V. x! {: s
unsigned char num4;
$ G2 k8 t# q9 p; G3 H% V unsigned char segR[10] ; //右- R7 d2 l, \4 u L5 F
unsigned char segL[10]; //左
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//左 5 @9 |, t( C/ S- v1 N8 o
segL[0] = 0x12;
1 e: B2 x# Z# ^2 x+ o# w segL[1] = 0xFA;
0 M" A2 ~# y! ?& \6 M segL[2] = 0x0E;
% \ ^) c6 T# U2 M( G/ v segL[3] = 0x2A;; K3 {" p- W. \4 G! m
segL[4] = 0xE2;/ i. ?7 X6 e- f% A
segL[5] = 0x63;5 b3 o0 {- n# c0 @9 D; Y
segL[6] = 0x2A;3 T- b# ~ X2 N- E
segL[7] = 0x7A;
! o. D: d: A( t( l& K/ B; {9 ^ segL[8] = 0x02;
8 _) w, X1 _" I0 `6 e9 z) l segL[9] = 0x22;
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# h3 ?& j' e, v1 U //右4 `- \/ b- ?4 _# E( B
segR[0] = 0x88;% W% L# M3 R( J+ c, d/ ^1 l
segR[1] = 0xEB;; Z; ]: H' m" G' c) i1 K$ g' ?0 U
segR[2] = 0x85;3 W2 I/ W' i& K. `8 V2 z# c5 b! A
segR[3] = 0x83;0 n& k8 g7 q" V; d' \
segR[4] = 0xE2;* S( T8 u. m& z/ v% \% j/ I
segR[5] = 0x92;
; q. V( q. M6 T/ T7 n# o9 E5 e segR[6] = 0x90;# S, I2 {. h3 |; f
segR[7] = 0xCB;+ H) P& [. _; {* E
segR[8] = 0x80;
( \# b0 F* F9 J: y segR[9] = 0x82;9 M5 f; c, b/ f2 ?+ B0 o! k; r0 [
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num1=num /1000;
) `9 p* I9 @6 j* |+ L" F# t num2=(num /100)%10;;' |# n% w7 t7 `7 }" d2 l
num3=(num /10)%100;;;
6 F7 p& Y2 b- k/ e) ]' j num4=num%10;
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1 Y( q( z2 }+ O$ i( s. ^5 R) s2 p, F dispA[0] = 0;
) _- l6 q5 r$ |& J8 x' G! q! H dispB[0] = 0;
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for (i=0;i<8;i++)
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dispA[i+1] = (segR[num2]>>i)&0x01;2 |; K% [, F2 C: c1 P
dispB[i+1] = (segR[num1]>>i)&0x01;
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7 F9 m" X4 c: u/ b( G, W dispA[i+9] = (segL[num4]>>i)&0x01;
' @1 N* r& k. |9 [6 W% ~! x dispB[i+9] = (segL[num3]>>i)&0x01;
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3 b9 d* r" `7 t7 @1 o# |# V# z8 m dispA[17] = 1;
1 a( t9 V& n( G8 ~) f dispB[17] = 0;2 T% \. }5 J9 w5 e
, q8 k. q. G2 }, @7 d dispA[18] = 0;# P- x0 e1 G: s: ~6 j" l
dispB[18] = 0;4 U8 E1 x) @- q) J
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/ R1 }2 ]( _. o4 q+ n6 ? dispB[19] = 0;6 j" q) q* y3 W! \
. z# g. i V& F/ @$ r yCLK = 1;
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for (i = 0;i<20;i++)# e7 \ ?" Q- y. m
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yCLK = 1;, _. \8 Z& K5 ~+ @4 w( ~, r
yData = dispA;+ A4 [& C- p" {, k4 i3 _
delay(2);
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+ w, m$ o, Y" X9 X( q$ k# R4 W {yDLEN=0;}
+ ~) \* `: U8 C+ i else
% D2 N/ R* n8 ` T2 m {yDLEN=1;}! c" @7 v H. K* T- ]' [/ X1 `
% \) ^( d4 p. c* l+ f delay(dailyCount);# V. `8 {; T# L+ }
yCLK = 0 ;4 W* c8 M* s: f+ P' Q9 R( o0 i4 V/ c
delay(dailyCount);% k- Z. O( r- h% ^
}
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, i; L, f7 l2 e! G8 ^! h yCLK=0;2 a2 g) Z: q6 J4 Z9 F# i Y" E
yData=0;
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delay(dailyCount*10);
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for (i = 0;i<20;i++)
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yCLK = 1;
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8 ~+ K8 c# m' [6 l) C. _ delay(dailyCount/10);
8 b( U- V" k, O if(i>17)
7 B2 h* b' w9 A& i4 S9 S. u/ o {yDLEN=0;}0 a0 Q: J1 T! ^+ @! }
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{yDLEN=1;}
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yCLK = 0 ;
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yCLK=0;
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yDLEN=0;
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void main(). W I! y; a. q, b
{
# P4 O: p: @. \. V$ t while(1)
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Q" l# _9 R+ ?3 X* i delay(4000);
1 x9 @% X! F; j5 J disp(0);: @7 t: L" n9 L6 m4 ~
}
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