D:\pythonPro\venv\Scripts\python.exe D:/pythonPro/Cohen-Sutherland.py Traceback (most recent call last): File "D:\pythonPro\Cohen-Sutherland.py", line 71, in <module> result = clip(x1, y1, x2, y2) File "D:\pythonPro\Cohen-Sutherland.py", line 36, in clip if code1 == 0 and code2 == 0: UnboundLocalError: local variable 'code1' referenced before assignment
时间: 2024-02-25 12:51:50 浏览: 100
非常抱歉,这是我犯的一个错误。在裁剪函数中,如果起点和终点都在裁剪区域内,则直接返回,但是在函数中没有定义code1和code2的值,因此会出现引用之前未定义的变量的错误。要解决这个问题,我们可以在函数开头添加以下代码,将code1和code2的初始值设为0:
```python
code1 = 0
code2 = 0
```
这样,即使起点和终点都在裁剪区域内,code1和code2也会被正确地初始化,就不会出现引用未定义变量的错误了。下面是修改后的完整代码:
```python
from PIL import Image, ImageDraw
# 定义裁剪区域的边界
x_min = 100
y_min = 100
x_max = 300
y_max = 300
# 定义线段的起点和终点
x1 = 50
y1 = 150
x2 = 350
y2 = 250
# 定义编码函数
def encode(x, y):
code = 0b0000
if x < x_min:
code |= 0b0001
elif x > x_max:
code |= 0b0010
if y < y_min:
code |= 0b0100
elif y > y_max:
code |= 0b1000
return code
# 定义裁剪函数
def clip(x1, y1, x2, y2):
code1 = 0
code2 = 0
while True:
# 如果起点和终点都在裁剪区域内,则直接返回
if code1 == 0 and code2 == 0:
return x1, y1, x2, y2
# 如果起点和终点都在裁剪区域外,则直接舍弃
elif (code1 & code2) != 0:
return None
# 否则,选取一个在裁剪区域外的点进行裁剪
else:
code = code1 if code1 != 0 else code2
if code & 0b0001:
x = x_min
y = y1 + (y2 - y1) * (x_min - x1) // (x2 - x1)
elif code & 0b0010:
x = x_max
y = y1 + (y2 - y1) * (x_max - x1) // (x2 - x1)
elif code & 0b0100:
y = y_min
x = x1 + (x2 - x1) * (y_min - y1) // (y2 - y1)
elif code & 0b1000:
y = y_max
x = x1 + (x2 - x1) * (y_max - y1) // (y2 - y1)
if code == code1:
x1, y1 = x, y
code1 = encode(x1, y1)
else:
x2, y2 = x, y
code2 = encode(x2, y2)
# 创建image对象和draw对象
img = Image.new('RGB', (400, 400), (255, 255, 255))
draw = ImageDraw.Draw(img)
# 绘制原始线段
draw.line((x1, y1, x2, y2), fill=(0, 0, 0), width=2)
# 调用裁剪函数进行裁剪
result = clip(x1, y1, x2, y2)
# 如果裁剪结果不为None,则绘制裁剪后的线段
if result is not None:
draw.line(result, fill=(255, 0, 0), width=2)
# 显示图片
img.show()
```
希望这次能够成功运行。
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