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单独的计时显示模块可以运行，单独的按键扫描模块也可以运行，就是凑起来的时候就出现问题。 看了好久没找出问题所在。#include
#define uchar unsigned char #define uint unsigned int #define ulong unsigned long ***it du = P2^0; ***it we = P2^1; ***it keyout0 = P3^0; ***it keylie0 = P3^4; ***it keylie1 = P3^5; ***it keylie2 = P3^6; ***it keylie3 = P3^7; ***it led = P1^0; uchar T0RL, T0RH; ulong Integer = 0; uchar decimal = 0; void ledscan(); void ledaccount(); void keyinit_1(); void keyscan(); void keydriver(); uchar code dutable[] = { 0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f}; uchar code wetable[] = { 0x7f,0xbf,0xdf,0xef,0xf7,0xfb,0xfd,0xfe}; uchar ledbuff[] = {0, 0, 0, 0, 0, 0}; uchar keystatic[] = {1,1,1,1}; bit stopkey = 0; bit keyinit = 0; void Timer_init(uint ms) { ulong tmp=0; TMOD = TMOD & 0XF0; TMOD = TMOD | 0X01; tmp = 11059200 / 12 ; tmp = tmp * ms / 1000; tmp = 65535 - tmp; T0RH = (uchar)(tmp >> 8); T0RL = (uchar)tmp; TH0 = T0RH; TL0 = T0RL; TR0 = 1; EA = 1; ET0 = 1; } void main() { Timer_init(2); keyout0 = 0; while(1) { if(keyinit == 1) { keyinit_1(); } keydriver(); } } void keydriver() { static uchar update[] ={1,1,1,1}; uchar i = 0; for(i=0; i<4; i++) { if(update[i] != keystatic[i]) { if(update[i] != 0) { if( i == 0) { stopkey = ~stopkey; } if(i == 1) { keyinit = 1; } } update[i] = keystatic[i]; } } } void keyscan() { uchar i = 0; static uchar keybuff[] = {0xff,0xff,0xff,0xff}; keybuff[0] = (keybuff[0]<<1) | keylie0; keybuff[1] = (keybuff[1]<<1) | keylie1; keybuff[2] = (keybuff[2]<<1) | keylie2; keybuff[3] = (keybuff[3]<<1) | keylie3; for(i=0; i<4; i++) { if(keybuff[i] == 0x00) { keystatic[i] = 0; } else if(keybuff[i] == 0xff) { keystatic[i] = 1; } } } void keyinit_1() { if(keyinit == 1) { Integer = 0; decimal = 0; stopkey = 0; keyinit = 0; } else keyinit = 0; } void ledscan() { static uchar i=0; P0 = 0x00; switch(i) { case 0: du = 1; P0=ledbuff[0]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 1: du = 1; P0=ledbuff[1]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 2: du = 1; P0=ledbuff[2]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 3: du = 1; P0=ledbuff[3]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 4: du = 1; P0=ledbuff[4]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 5: du = 1; P0=ledbuff[5]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 6: du = 1; P0=ledbuff[6]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i++; break; case 7: du = 1; P0=ledbuff[7]; du = 0; P0 = 0XFF; we = 1; P0 = wetable[i]; we = 0; i=0; break; default : break; } } void ledaccount() { if(stopkey == 0) { decimal++; ledbuff[0] = dutable[(decimal%10)]; ledbuff[1] = dutable[(decimal/10)]; if(decimal>=100) { decimal=0; Integer++; ledbuff[2] = dutable[(Integer%10 )]|0x80; ledbuff[3] = dutable[(Integer/10%10)]; ledbuff[4] = dutable[(Integer/100%10)]; ledbuff[5] = dutable[(Integer/1000%10 )]; ledbuff[6] = dutable[(Integer/10000%10 )]; ledbuff[7] = dutable[(Integer/100000%10 )]; if(Integer == 100000) { Integer = 0; } } } } void interrupTimer0() interrupt 1 { static uchar mun=0; TH0 = T0RH; TL0 = T0RL; mun++; ledscan(); if(mun == 10) { mun =0 ; ledaccount(); } keyscan(); } 
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5个回答
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回帖奖励 +1 分积分
你也没有说出是什么问题呀，
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实现的功能其实很简单，就是有两个按键，一个是停止按键按下去就要停止，一个是复位清零。。。可是程序会跑飞 |
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这程序是你自己写的吗？如果是我写的话就不会写的这么复杂，大概思路是：程序一开始就计数，用定时器1或者0都可以，然后再定时中断服务程序里设定终值（要看使用多少位数码管，2位的话就是99,3位就是999）等计数到了种植就清零从头开始计数！在主函数里不停的扫描按键函数（比如设定key1为停止键，key2为清零键）当按下key1时，只要使TR0=0就可以了！当按下key2时，只要使num=0就OK了（num是计数参数）！
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ningandro 发表于 2015-11-1 19:23 这个问题我已经解决了，谢谢。其实如果按照你的说法不是说不行。只不过可能程序的移植性没那么好。。 |
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