In function ‘init_lcd’:
时间: 2024-04-27 22:22:42 浏览: 15
这个提示很可能是一个函数定义或声明的问题。`init_lcd`函数可能没有正确定义或声明,或者在定义或声明之前调用了该函数。请确保:
1. `init_lcd`函数已经正确声明或定义,如果您是在别的文件中定义该函数,请确保在当前文件中包含了该函数的头文件。
2. 如果您在当前文件中定义`init_lcd`函数,请确保该函数在使用之前已经定义或声明。
3. 如果您在调用`init_lcd`函数之前定义或声明了该函数,请确保您使用的函数名与定义或声明中的函数名一致。同时,注意函数的参数类型和数量是否与定义或声明中的一致。
相关问题
/tmp/ccOXlJri.o: In function `task2': project.c:(.text+0x661): undefined reference to `lcd_open' project.c:(.text+0x666): undefined reference to `mmap_lcd' project.c:(.text+0x672): undefined reference to `linux_v4l2_yuyv_init' project.c:(.text+0x67c): undefined reference to `linux_v4l2_start_yuyv_capturing' project.c:(.text+0x68b): undefined reference to `linux_v4l2_get_yuyv_data' project.c:(.text+0x6ab): undefined reference to `show_video_data' /tmp/ccOXlJri.o: In function `task3': project.c:(.text+0x6d7): undefined reference to `lcd_open' project.c:(.text+0x6dc): undefined reference to `mmap_lcd' project.c:(.text+0x724): undefined reference to `lcd_close' collect2: error: ld returned 1 exit status
这是一个编译错误,表明在链接时找不到lcd_open、mmap_lcd、linux_v4l2_yuyv_init、linux_v4l2_start_yuyv_capturing、linux_v4l2_get_yuyv_data和show_video_data这些函数的定义。这可能是因为您没有正确地链接相应的库文件或源文件。请检查您的编译命令,确保正确地包含了这些函数所在的库文件或源文件。
import RPi.GPIO as GPIO from LCD1602 import LCD_1602 import time BtnPin = 13 R = 4 G = 12 B = 6 TRIG = 17 ECHO = 18 flag = 2 buzzer = 20 GPIO.setwarnings(False) GPIO.setmode(GPIO.BCM) GPIO.setup(TRIG,GPIO.OUT,initial=GPIO.LOW) GPIO.setup(ECHO,GPIO.IN) GPIO.setup(R,GPIO.OUT) GPIO.setup(B, GPIO.OUT) GPIO.setup(G, GPIO.OUT) GPIO.setup(buzzer, GPIO.OUT) GPIO.setup(BtnPin,GPIO.IN, pull_up_down=GPIO.PUD_UP) GPIO.output(buzzer,GPIO.HIGH) m_lcd = LCD_1602(Address=0x27, bus_id=1, bl=1) flag = m_lcd.lcd_init() while True: if GPIO.input(BtnPin) == 0: flag += 1 elif GPIO.input(BtnPin) == 1: pass if flag % 2 == 0: for i in range(10): GPIO.output(G,GPIO.LOW) GPIO.output(B,GPIO.LOW) GPIO.output(buzzer,GPIO.HIGH) GPIO.output(TRIG,GPIO.HIGH) time.sleep(0.000015) GPIO.output(TRIG,GPIO.LOW) while not GPIO.input(ECHO): pass t1 = time.time() while GPIO.input(ECHO): pass t2 = time.time() distance = round((t2-t1) * 340 / 2,5) if distance < 0.2: GPIO.output(B,GPIO.HIGH) GPIO.output(buzzer,GPIO.LOW) print("当前的距离为:",distance) a = '%f'%distance m_lcd.lcd_display_string(0, 0, 'The distance is') m_lcd.lcd_display_string(0, 1, a) m_lcd.lcd_display_string(8, 1, 'm') time.sleep(1) elif flag % 2 == 1: GPIO.output(G,GPIO.HIGH) GPIO.cleanup(),帮我封装一下代码
Sure, here's the code wrapped in a function:
```python
import RPi.GPIO as GPIO
from LCD1602 import LCD_1602
import time
def run_distance_sensor():
BtnPin = 13
R = 4
G = 12
B = 6
TRIG = 17
ECHO = 18
flag = 2
buzzer = 20
GPIO.setwarnings(False)
GPIO.setmode(GPIO.BCM)
GPIO.setup(TRIG,GPIO.OUT,initial=GPIO.LOW)
GPIO.setup(ECHO,GPIO.IN)
GPIO.setup(R,GPIO.OUT)
GPIO.setup(B, GPIO.OUT)
GPIO.setup(G, GPIO.OUT)
GPIO.setup(buzzer, GPIO.OUT)
GPIO.setup(BtnPin,GPIO.IN, pull_up_down=GPIO.PUD_UP)
GPIO.output(buzzer,GPIO.HIGH)
m_lcd = LCD_1602(Address=0x27, bus_id=1, bl=1)
flag = m_lcd.lcd_init()
while True:
if GPIO.input(BtnPin) == 0:
flag += 1
elif GPIO.input(BtnPin) == 1:
pass
if flag % 2 == 0:
for i in range(10):
GPIO.output(G,GPIO.LOW)
GPIO.output(B,GPIO.LOW)
GPIO.output(buzzer,GPIO.HIGH)
GPIO.output(TRIG,GPIO.HIGH)
time.sleep(0.000015)
GPIO.output(TRIG,GPIO.LOW)
while not GPIO.input(ECHO):
pass
t1 = time.time()
while GPIO.input(ECHO):
pass
t2 = time.time()
distance = round((t2-t1) * 340 / 2,5)
if distance < 0.2:
GPIO.output(B,GPIO.HIGH)
GPIO.output(buzzer,GPIO.LOW)
print("当前的距离为:",distance)
a = '%f'%distance
m_lcd.lcd_display_string(0, 0, 'The distance is')
m_lcd.lcd_display_string(0, 1, a)
m_lcd.lcd_display_string(8, 1, 'm')
time.sleep(1)
elif flag % 2 == 1:
GPIO.output(G,GPIO.HIGH)
GPIO.cleanup()
```
You can call the `run_distance_sensor()` function to start running the distance sensor code.
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