请设计一段代码适用于FS-STM32F407单片机开发平台,构造一个密码验证系统,如果输入的密码错误蜂鸣器响
时间: 2024-05-01 19:23:50 浏览: 142
以下是一个简单的密码验证系统,使用了FS-STM32F407单片机开发平台和一个蜂鸣器模块。代码中默认密码为 "1234",如果用户输入的密码不正确,蜂鸣器将响一次。
```
#include "stm32f4xx.h"
#define PASSWORD "1234"
void delay(uint32_t count) {
for(uint32_t i = 0; i < count; i++) {
__NOP();
}
}
int main(void) {
RCC->AHB1ENR |= RCC_AHB1ENR_GPIODEN;
GPIOD->MODER |= GPIO_MODER_MODE12_0;
GPIOD->OTYPER &= ~GPIO_OTYPER_OT12;
GPIOD->PUPDR &= ~GPIO_PUPDR_PUPD12;
GPIOD->MODER |= GPIO_MODER_MODE13_0;
GPIOD->OTYPER &= ~GPIO_OTYPER_OT13;
GPIOD->PUPDR &= ~GPIO_PUPDR_PUPD13;
GPIOD->MODER |= GPIO_MODER_MODE14_0;
GPIOD->OTYPER &= ~GPIO_OTYPER_OT14;
GPIOD->PUPDR &= ~GPIO_PUPDR_PUPD14;
GPIOD->MODER |= GPIO_MODER_MODE15_0;
GPIOD->OTYPER &= ~GPIO_OTYPER_OT15;
GPIOD->PUPDR &= ~GPIO_PUPDR_PUPD15;
while(1) {
// Get user input for password
char input[5];
for(int i = 0; i < 4; i++) {
// Wait for user input
while(!(USART2->SR & USART_SR_RXNE));
char c = (char)USART2->DR;
input[i] = c;
USART2->DR = c;
while(!(USART2->SR & USART_SR_TC));
}
input[4] = '\0';
// Check if password is correct
if(strcmp(input, PASSWORD) == 0) {
// Password is correct
GPIOD->ODR |= GPIO_ODR_OD12;
delay(1000000);
GPIOD->ODR &= ~GPIO_ODR_OD12;
} else {
// Password is incorrect
GPIOD->ODR |= GPIO_ODR_OD13;
delay(1000000);
GPIOD->ODR &= ~GPIO_ODR_OD13;
// Beep the buzzer once
GPIOD->ODR |= GPIO_ODR_OD14;
delay(50000);
GPIOD->ODR &= ~GPIO_ODR_OD14;
}
}
}
```
代码的主要思路是通过串口从用户那里获取密码输入。首先,我们需要在代码中定义一个默认的密码(在本例中为 "1234")。然后,我们使用STM32的串口(USART2)获取用户输入的密码。我们等待用户输入四个字符,然后将它们存储在一个字符数组中。接下来,我们将输入的密码与默认密码进行比较。如果密码正确,我们将点亮绿色LED灯1秒钟。如果密码不正确,我们将点亮红色LED灯并发出一声蜂鸣器。
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