通过Tkinter中的Button将参数传递给函数，循环中出现starnge行为

我目前正在尝试修复python书“ Python下一步”中的一个错误，该错误作者尚未修复，并在代码中留下了注释：“稍后修复”

我的第一个解决方案失败了，但是第二个解决方案通过删除了循环成功了。问题是我不知道为什么第一个解决方案失败了！

解决方案1：

当用户使用循环单击Tkinter制造的计算器中的按钮时，Button对象和称为click的函数仅从lambda参数打印大写字母C.在谈论。

解决方案2：

删除生成按钮的循环并重复手动编码每个按钮！这可以正常工作，正如Brian kernighan所建议的那样：首先使它工作，然后再提高效率！

代码：

# myCalculator3_final.py

from Tkinter import *
from decimal import *

# key press function:
def click(key):
        display.insert(END, key)


##### main:
window = Tk()
window.title("MyCalculator")

# create top_row frame
top_row = Frame(window)
top_row.grid(row=0, column=0, columnspan=2, sticky=N)

# use Entry widget for an editable display
display = Entry(top_row, width=45, bg="light green")
display.grid()

# create num_pad_frame
num_pad = Frame(window)
num_pad.grid(row=1, column=0, sticky=W)

# This method of passing an argument to click work! Loop removed and buttons hand code
#>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
#-------------------------------------------------------------------------
# create num_pad buttons passing an argument to the command function click
#-------------------------------------------------------------------------   
seven = Button(num_pad, text="7", width=5, command=lambda :click("7"))
seven.grid(row=0,column=0)
eight = Button(num_pad, text="8", width=5, command=lambda :click("8"))
eight.grid(row=0,column=1)
nine= Button(num_pad, text="9", width=5, command=lambda :click("9"))
nine.grid(row=0,column=2)
four= Button(num_pad, text="4", width=5, command=lambda :click("4"))
four.grid(row=1,column=0)
five= Button(num_pad, text="5", width=5, command=lambda :click("5"))
five.grid(row=1,column=1)
six= Button(num_pad, text="6", width=5, command=lambda :click("6"))
six.grid(row=1,column=2)
one= Button(num_pad, text="1", width=5, command=lambda :click("1"))
one.grid(row=2,column=0)
two= Button(num_pad, text="2", width=5, command=lambda :click("2"))
two.grid(row=2,column=1)
three= Button(num_pad, text="3", width=5, command=lambda :click("3"))
three.grid(row=2,column=2)
zero= Button(num_pad, text="0", width=5, command=lambda :click("0"))
zero.grid(row=2,column=0)
#---------------------------------------------------------------------------   



# calculate the row, column for button

# create operator_frame
operator = Frame(window)
operator.grid(row=1, column=1, sticky=E)

operator_list = [
'*', '/',  
'+', '-',
'(', ')',
'C' ]

# The authors code and I have added  the same lambda function as above but 
#it just prints out capital C
#>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
r = 0
c = 0
for btn_text in operator_list:
    Button(operator, text=btn_text, width=5, command=lambda: click(btn_text)).grid(row=r,column=c)
    c = c+1
    if c > 1:
        c = 0
        r = r+1


##### Run mainloop
window.mainloop()

问题：

为什么单击lambda调用方法以传递参数无法在循环中起作用，而只显示C，但是如果我删除循环，则它起作用！