我最近开发了一个Turtle程序来接收指令（转弯，平板，下降，圆圈等），并且工作正常。但是，当我尝试添加一定次数迭代一定数量的步骤的能力时，Turtle程序将运行一次迭代，然后由于未知原因而无法响应。我认为这可能是因为'@'是程序迭代所必需的，但没有任何东西真的有效。
该程序的输入语法是：
F ### - 转发“###”长度
B ### - 向后为“###”长度
L ### - 向左转“###”度
R ### - 向右转“###”度
C ### - 绘制一个给定（###）半径的圆圈
你 - 拿起笔
D - 将笔放下
我＃... @ - 迭代指令块“＃”次数
C250 I3 F050 R180 C225 @ B100 C125
CODE：
import turtle
def evaluate(commands):
"""
It's basically supposed to work
:param commands:
:return: None
"""
counter = 0
commands_length = len(commands)
while counter < commands_length:
# If the letter U is encountered, the turtle goes up
# counter increases by 2 to parse for the next character
if commands[counter] == 'U':
turtle.up()
print('up')
counter += 2
# If the letter D is encountered, the turtle goes down
# counter increases by 2 to parse for the next character
elif commands[counter] == 'D':
turtle.down()
print('down')
counter += 2
# If the letter F is encountered, the turtle moves forward
# by the amount denoted by the three following numbers
elif commands[counter] == 'F':
turtle.forward(int(commands[counter + 1: counter + 4: 1]))
print('forward(' + commands[counter + 1: counter + 4: 1] + ')')
counter += 5
# If the letter C is encountered, the turtle draws a circle
# with radius denoted by the three following numbers
elif commands[counter] == 'C':
turtle.circle(int(commands[counter + 1: counter + 4: 1]))
print('circle(' + (commands[counter + 1: counter + 4: 1]) + ')')
counter += 5
# if the letter B is encountered, the turtle moves backward
# by the amount denoted by the three following numbers
elif commands[counter] == 'B':
turtle.backward(int(commands[counter + 1: counter + 4: 1]))
print('backward(' + (commands[counter + 1: counter + 4: 1]) + '}')
counter += 5
# if the letter L is encountered, the turtle turns to its left
# by the angle denoted by the three following numbers
elif commands[counter] == 'L':
turtle.left(int(commands[counter + 1: counter + 4: 1]))
print('left(' + (commands[counter + 1: counter + 4: 1]) + ')')
counter += 5
# if the letter R is encountered, the turtle turns to its right
# by the angle denoted by the three following numbers
elif commands[counter] == 'R':
turtle.right(int(commands[counter + 1: counter + 4: 1]))
print('right(' + (commands[counter + 1: counter + 4: 1]) + ')')
counter += 5
elif commands[counter] == 'I':
counter += 3
loop = commands[counter: commands.index("@") + 1]
loop_counter = 0
loop_length = len(loop)
while loop_counter < loop_length:
for _ in range(counter + (commands.index("@") - 1)):
for x in range(loop.index('@')):
# If the letter U is encountered, the turtle goes up
# counter increases by 2 to parse for the next character
if loop[loop_counter] == 'U':
turtle.up()
print('up')
loop_counter += 2
# If the letter D is encountered, the turtle goes down
# counter increases by 2 to parse for the next character
elif loop[loop_counter] == 'D':
turtle.down()
print('down')
loop_counter += 2
# If the letter F is encountered, the turtle moves forward
# by the amount denoted by the three following numbers
elif loop[loop_counter] == 'F':
turtle.forward(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('forward(' + loop[loop_counter + 1: loop_counter + 4: 1] + ')')
loop_counter += 5
# If the letter C is encountered, the turtle draws a circle
# with radius denoted by the three following numbers
elif loop[loop_counter] == 'C':
turtle.circle(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('circle(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
# if the letter B is encountered, the turtle moves backward
# by the amount denoted by the three following numbers
elif loop[loop_counter] == 'B':
turtle.backward(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('backward(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + '}')
loop_counter += 5
# if the letter L is encountered, the turtle turns to its left
# by the angle denoted by the three following numbers
elif loop[loop_counter] == 'L':
turtle.left(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('left(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
# if the letter R is encountered, the turtle turns to its right
# by the angle denoted by the three following numbers
elif loop[loop_counter] == 'R':
turtle.right(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('right(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
turtle.done()
def main() -> None:
user_input = input("Enter Commands:")
evaluate(user_input.upper())
turtle.mainloop()
if __name__ == '__main__':
main()
答案 0 :(得分：1)
我认为你的主要问题是这句话：commands
这从commands.index("@", counter)开始搜索，所以如果你有两个循环，第二个会找到错误的终结符。您需要从当前计数器开始搜索：turtle.done()
我也不相信你的重复代码是正确的。您对turtle.mainloop()的来电是可疑的，因为I是多余的，所以我不予理会。
下面只修改了您的I4 F050 L090 @ I6 L060 F030 @处理程序代码。您可以使用小型双循环程序对其进行测试：import turtle
def evaluate(commands):
"""
It's basically supposed to work
:param commands:
:return: None
"""
counter = 0
commands_length = len(commands)
while counter < commands_length:
# ...
if commands[counter] == 'I':
repeat = int(commands[counter + 1: counter + 2: 1])
counter += 3
loop_end = commands.index("@", counter)
loop = commands[counter: loop_end]
counter = loop_end + 2
loop_length = len(loop)
for _ in range(repeat):
loop_counter = 0
while loop_counter < loop_length:
# If the letter U is encountered, the turtle goes up
# counter increases by 2 to parse for the next character
if loop[loop_counter] == 'U':
turtle.up()
print('up')
loop_counter += 2
# If the letter D is encountered, the turtle goes down
# counter increases by 2 to parse for the next character
elif loop[loop_counter] == 'D':
turtle.down()
print('down')
loop_counter += 2
# If the letter F is encountered, the turtle moves forward
# by the amount denoted by the three following numbers
elif loop[loop_counter] == 'F':
turtle.forward(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('forward(' + loop[loop_counter + 1: loop_counter + 4: 1] + ')')
loop_counter += 5
# If the letter C is encountered, the turtle draws a circle
# with radius denoted by the three following numbers
elif loop[loop_counter] == 'C':
turtle.circle(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('circle(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
# if the letter B is encountered, the turtle moves backward
# by the amount denoted by the three following numbers
elif loop[loop_counter] == 'B':
turtle.backward(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('backward(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + '}')
loop_counter += 5
# if the letter L is encountered, the turtle turns to its left
# by the angle denoted by the three following numbers
elif loop[loop_counter] == 'L':
turtle.left(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('left(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
# if the letter R is encountered, the turtle turns to its right
# by the angle denoted by the three following numbers
elif loop[loop_counter] == 'R':
turtle.right(int(loop[loop_counter + 1: loop_counter + 4: 1]))
print('right(' + (loop[loop_counter + 1: loop_counter + 4: 1]) + ')')
loop_counter += 5
def main() -> None:
user_input = input("Enter Commands:")
evaluate(user_input.upper())
turtle.mainloop()
if __name__ == '__main__':
main()
I我认为一个更好的解决方案，而不是重新实现evaluate()处理程序的所有内容，将设计I，以便 <img ng-repeat="imagePath in user.imagePath" ng-src="{{imagePath }}">
处理程序可以递归地调用它来执行环路。