#include <reg51.h> 1 2 3 sbit button=p1^0; 4 sbit led=p1^1; 5 bit led state=0; 6 7 8日void delay(unsigned int ms){ 9 unsigned int i,j; 10 for(i=ms;i>0;i--) 11 for(j=110;j>0;j--); 1213 14 void main()【 15 while(1){ 16 if (button==0){ 17 delay(20); 18 if (button==0){ 19 1ed=0; 20 21 } 做注释
我可以帮助你为这段代码添加注释。以下是带有详细注释的代码: ```c #include <reg51.h> // 包含8051单片机寄存器定义的头文件 // 定义特殊功能位(SFR) sbit button = P1^0; // 将P1口的第0位定义为button按钮输入 sbit led = P1^1; // 将P1口的第1位定义为led输出 // 定义一个全局变量来保存LED的状态 bit led_state = 0; // 延时函数,参数ms表示延时的毫秒数 void delay(unsigned int ms) { unsigned int i, j; // 定义两个无符号整型变量i和j for (i = ms; i > 0; i--) // 外层循环控制延时的毫秒数 for (j = 110; j > 0; j--); // 内层循环控制每次延时的微秒数 } // 主函数 void main() { while (1) { // 无限循环 if (button == 0) { // 如果button按下(假设低电平有效) delay(20); // 延时20毫秒,用于消抖 if (button == 0) { // 再次检查button是否仍然按下 led = ~led_state; // 切换LED状态 led_state = ~led_state; // 更新led_state的值 } } } } ``` 解释: 1. `#include <reg51.h>`: 包含8051单片机的寄存器定义。 2. `sbit button = P1^0;`: 将P1口的第0位定义为button按钮输入。 3. `sbit led = P1^1;`: 将P1口的第1位定义为led输出。 4. `bit led_state = 0;`: 定义一个全局变量`led_state`来保存LED的状态。 5. `void delay(unsigned int ms)`: 延时函数,通过嵌套循环实现延时。 6. `void main()`: 主函数,程序从这里开始执行。 7. `while (1)`: 无限循环,使程序一直运行。 8. `if (button == 0)`: 检查button是否按下。 9. `delay(20);`: 延时20毫秒,用于消除按键抖动。 10. `if (button == 0)`: 再次检查button是否仍然按下。 11. `led = ~led_state;`: 切换LED状态。 12. `led_state = ~led_state;`: 更新`led_state`的值。 希望这些注释能帮助你更好地理解这段代码的功能和工作原理。 [2024-10-29 15:31:00 | AI写代码神器 | 693点数解答]
- [讯飞星火] #include <reg51.h> 3 sbit button=p1^0; 4 sbit led=pl^1; 5 6 bit led_state=0; 7 8日void delay(unsigned int ms){ 9 unsigned int i,j: 10 for(i=ms;i>0;i--) 11 for(j=110;j>0;j--); 12 } 13 14日void main(){ 15 while(1){ 16 白 if (button==0){ 17 delay(20); 18 if (button-=0){ 19 while (button==ah; 20 led state=~led state; 21 led=led state; 22 } 写详细注释(697点数解答 | 2024-10-29 14:54:31)263
- [讯飞星火] #include <reg51.h> 1 2 3 sbit button=p1^0; 4 sbit led=p1^1; 5 bit led state=0; 6 7 8日void delay(unsigned int ms){ 9 unsigned int i,j; 10 for(i=ms;i>0;i--) 11 for(j=110;j>0;j--); 1213 14 void main()【 15 while(1){ 16 if (button==0){ 17 delay(20); 18 if (button==0){ 19 1ed=0; 20 21 } 做详细注释(830点数解答 | 2024-10-29 15:29:41)276
- [讯飞星火] #include <reg51.h> 1 2 3 sbit button=p1^0; 4 sbit led=p1^1; 5 bit led state=0; 6 7 8日void delay(unsigned int ms){ 9 unsigned int i,j; 10 for(i=ms;i>0;i--) 11 for(j=110;j>0;j--); 1213 14 void main()【 15 while(1){ 16 if (button==0){ 17 delay(20); 18 if (button==0){ 19 1ed=0; 20 21 } 做注释(693点数解答 | 2024-10-29 15:31:00)398
- [讯飞星火] #include <reg51.h> 1 2 3 sbit button=p1^0; 4 sbit led=p1^1; 5 bit led state=0; 6 7 8日void delay(unsigned int ms){ 9 unsigned int i,j; 10 for(i=ms;i>0;i--) 11 for(j=110;j>0;j--); 1213 14 void main()【 15 while(1){ 16 if (button==0){ 17 delay(20); 18 if (button==0){ 19 1ed=0; 20 21 }(570点数解答 | 2024-10-29 15:32:29)243
- [DeepSeek] #include <reg51.h> sbit LED = P1^0; sbit KEY = P1^1; void delay(unsigned int ms) { unsigned int i; for(; ms>0; ms--) for(i=0; i<120; i++); } void main() { while(1) { if(KEY == 0) { delay(20); if(KEY == 0) { LED = ~LED; while(KEY == 0); } } } } (326点数解答 | 2025-06-22 23:19:40)187
- [阿里通义] 把这个代码修改正确#include<reg52.h> sbit key = p3^0; // 按键连接到p3.0 unsigned char code seg_code[] = {0x3f, 0x06, 0x5b, 0x4f, 0x66, 0x6d, 0x7d, 0x07, 0x7f, 0x6f}; // 数码管段码 unsigned char code dash_code = 0x40; // 横杠段码 unsigned int second =0; // 秒表计数 bit running = 0; // 秒表运行状态 void timer0_isr() interrupt 1 { static unsigned int count =0; th0 = (65536 - 1000) / 256; tl0 = (65536 - 1000) %256; count++; if (count >= 1000) { count = 0; if (running) { second++; } } } void delay(unsigned int ms) { unsigned int i(776点数解答 | 2024-10-23 22:51:51)272
- [DeepSeek] time_space_table: /1213/sample.in:MLE mem=196608k time=158ms /1213/star0.in:MLE mem=196608k time=4ms /1213/star1.in:MLE mem=196608k time=4ms /1213/star2.in:MLE mem=196608k time=3ms /1213/star3.in:MLE mem=196608k time=2ms /1213/star4.in:MLE mem=196608k time=4ms /1213/star5.in:MLE mem=196608k time=2ms /1213/star6.in:MLE mem=196608k time=3ms /1213/star7.in:MLE mem=196608k time=3ms 上题内存超限,重写(714点数解答 | 2026-01-06 17:49:47)22
- [字节豆包] #include<iostream> #include<complex.h> #include <complex> #include <math.h> using namespace std; #define pi 3.14159265358979323846 int main() { file* stream; stream = fopen("4000-red.raw","rb"); unsigned char* image = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); fread(image,4000*4000,sizeof(unsigned char),stream); unsigned char* image2 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); unsigned char* image3 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned(413点数解答 | 2024-11-02 18:16:35)300
- [DeepSeek] 外部中断 0 由边沿触发,控制 P2 口的 8 个发光二极管依次点亮。C51 程序如下: #include <reg51.h> void ex0_isr() interrupt 0 { static unsigned char led = 0x01; P2 = led; led = (led << 1) | (led >> 7); } void main() { IT0 = 1; EX0 = 1; EA = 1; P2 = 0x01; while(1); } (228点数解答 | 2025-06-22 23:18:42)139
- [DeepSeek] #include <reg51.h> void ex0_isr() interrupt 0 { static unsigned char led = 0x01; P2 = led; led = (led << 1) | (led >> 7); } void main() { IT0 = 1; EX0 = 1; EA = 1; P2 = 0x01; while(1); }(214点数解答 | 2025-06-22 23:19:09)168
- [字节豆包] 采用c++语言,实现如下功能: 实现古典密码中的移位密码和仿射密码,具体实现如下接口: c++接口: <<<<<移位密码>>>>> bool is_valid_s(unsinged char k) { 判断k是否为合理的密钥 } int encrypt_s(unsigned char* p, unsigned char k) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_s(unsigned char* c, unsigned char k) { 密钥合法则返回1,且明文覆盖密文: 密钥不合法则返回0. } <<<<<仿射密码>>>>> bool in_valid_a(unsinged char a, unsigned char b) { 判断a,b是否为合理的密钥 } int encrypt_a(unsigned char* p, unsigned char a, unsigned char b) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_a(unsigned char* c, unsign(812点数解答 | 2024-12-18 16:02:36)331
- [讯飞星火] #include <stc15f2k60s2.h> // 数码管段选码 unsigned char code seg_code[] = { 0xc0, 0xf9, 0xa4, 0xb0, 0x99, 0x92, 0x82, 0xf8, 0x80, 0x90, 0x88, 0x83, 0xc6, 0xa1, 0x86, 0x8e }; // 数码管位选码 unsigned char code bit_code[] = { 0xfe, 0xfd, 0xfb, 0xf7, 0xef, 0xdf, 0xbf, 0x7f }; void delay(unsigned int t) { while (t--) ; } void initadc() { p1asf = 0xff; // 将 p1 口设置为模拟输入口 adc_contr = 0x80; // 打开 adc 电源 delay(2); // 适当延时等待 adc 电源稳定 } unsigned int getadcresult() { adc_con(939点数解答 | 2024-11-07 17:31:31)241