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如何用 C++ 模拟一个点阵显示器

news 2026/6/10 22:34:43

之前在外面看到的一个大一学生学习 C 语言，学到二维数组和多维数组那块的一个课后作业。

要求是利用控制台的等宽字体模拟一个5×75\times 75×7的点阵字体（其实就是单片机用的那种液晶屏）并按照这种方式输出系统时间。

（图片来源网络，侵删）

Solution

纯模拟，算法层面完全没什么难的，但是调试非常考验耐心 👇

如何对齐字符，如何保证输出不错乱都需要仔细调整；
控制台窗口宽度不够，同样会导致输出混乱；
打表 - 最枯燥的过程，没有之一。

参考实现

#if_MSVC_LANG#pragmawarning(disable:4996)#endif#include<windows.h>#include<time.h>#include<stdio.h>#if!__cplusplus#defineconstexprconst#defineboolint#endif// 枚举10个数码constexprboolnum[10][7][5]={{{0,1,1,1,0},{1,0,0,0,1},{1,0,0,1,1},{1,0,1,0,1},{1,1,0,0,1},{1,0,0,0,1},{0,1,1,1,0}},{{0,0,1,0,0},{0,1,1,0,0},{0,0,1,0,0},{0,0,1,0,0},{0,0,1,0,0},{0,0,1,0,0},{0,1,1,1,0}},{{0,1,1,1,0},{1,0,0,0,1},{0,0,0,0,1},{0,0,0,1,0},{0,0,1,0,0},{0,1,0,0,0},{1,1,1,1,1}},{{0,1,1,1,0},{1,0,0,0,1},{0,0,0,0,1},{0,0,1,1,0},{0,0,0,0,1},{1,0,0,0,1},{0,1,1,1,0}},{{0,0,0,1,0},{0,0,1,1,0},{0,1,0,1,0},{1,0,0,1,0},{1,1,1,1,1},{0,0,0,1,0},{0,0,0,1,0}},{{1,1,1,1,1},{1,0,0,0,0},{1,1,1,1,0},{0,0,0,0,1},{0,0,0,0,1},{1,0,0,0,1},{0,1,1,1,0}},{{0,0,1,1,0},{0,1,0,0,0},{1,0,0,0,0},{1,1,1,1,0},{1,0,0,0,1},{1,0,0,0,1},{0,1,1,1,0}},{{1,1,1,1,1},{0,0,0,0,1},{0,0,0,1,0},{0,0,1,0,0},{0,1,0,0,0},{0,1,0,0,0},{0,1,0,0,0}},{{0,1,1,1,0},{1,0,0,0,1},{1,0,0,0,1},{0,1,1,1,0},{1,0,0,0,1},{1,0,0,0,1},{0,1,1,1,0}},{{0,1,1,1,0},{1,0,0,0,1},{1,0,0,0,1},{0,1,1,1,1},{0,0,0,0,1},{0,0,0,1,0},{0,1,1,0,0}}};// 用bool数组表示六个时间数码以及两个冒号, 每两个字符之间有一个像素宽度的间隔, 因此6个数字加上两个冒号共8个字符有7个间隔, 需要一个7行47列的数组booltime_display[7][48];booldate_display[7][64];// 每个数字的位置constexprinttime_position[6]={0,6,18,24,36,42},date_position[8]={0,6,12,18,30,36,48,54};// 用中括号填充数字, xstart表示起始列(只可为0,6,18,24,36,42), number表示要填充的数字(-1表示冒号)voidinsert_time(intxstart,intdigit){for(inti=0;i<=6;i+=1)for(intj=xstart;j<=xstart+4;j+=1)time_display[i][j]=num[digit][i][j-xstart];}voidinsert_date(intxstart,intdigit){for(inti=0;i<=6;i+=1)for(intj=xstart;j<=xstart+4;j+=1)date_display[i][j]=num[digit][i][j-xstart];}// 初始化操作voidinit(void){// 插入冒号time_display[1][14]=time_display[5][14]=time_display[1][32]=time_display[5][32]=1;date_display[6][26]=date_display[6][25]=date_display[5][26]=date_display[5][25]=date_display[6][43]=date_display[6][44]=date_display[5][43]=date_display[5][44]=1;// system("color ce");}// 获取时间填充数字voidinput_time(void){time_t tp;time(&tp);tm*p=localtime(&tp);intdigit[6]={p->tm_hour/10,p->tm_hour%10,p->tm_min/10,p->tm_min%10,p->tm_sec/10,p->tm_sec%10};intdate[8]={(p->tm_year+1900)/1000,((p->tm_year+1900)%1000)/100,((p->tm_year+1900)%100)/10,(p->tm_year+1900)%10,(p->tm_mon+1)/10,(p->tm_mon+1)%10,p->tm_mday/10,p->tm_mday%10};for(inti=0;i<=5;i+=1)insert_time(time_position[i],digit[i]);for(inti=0;i<=7;i+=1)insert_date(date_position[i],date[i]);}// 输出与清屏voidprint_time(void){system("cls");printf("\n");for(inti=0;i<=6;i+=1){printf(" ");for(intj=0;j<=59;j+=1)printf(date_display[i][j]?#if_WIN32_WINNT==_WIN32_WINNT_WIN10"\u2588\u2588"#elif_WIN32_WINNT==_WIN32_WINNT_WIN7"\u2588"#elif_WIN32_WINNT==_WIN32_WINNT_WINXP"[]"#endif:" ");putchar(10);}printf("\n\n");for(inti=0;i<=6;i+=1){printf(" ");for(intj=0;j<=46;j+=1)printf(time_display[i][j]?#if_WIN32_WINNT==_WIN32_WINNT_WIN10"\u2588\u2588"#elif_WIN32_WINNT==_WIN32_WINNT_WIN7"\u2588"#elif_WIN32_WINNT==_WIN32_WINNT_WINXP"[]"#endif:" ");putchar(10);}Sleep(950);}intmain(){init();while(1){input_time();print_time();}}