Python函数第三节

# 3.定义一个方法list_info(list), 参数list为列表对象，怎么保证在函数里对列表list进行一些相关的操作，
# 不会影响到原来列表的元素值，比如：a = [1,2,3]

a = [1, 2, 3]

def list_info(list_m):
temp = [list_m[i] for i in range(0, len(list_m))]
temp[0] = 5
return temp
print a
print list_info(a)
print a

[1, 2, 3]
[5, 2, 3]
[1, 2, 3]

在这里面需要注意的是，函数里面直接写temp=list_m的话在改变temp的过程中，会改变list_m的值
a = [1, 2, 3]

def list_info(list_m):
temp = list_m
temp[0] = 5
return temp
print a
print list_info(a)
print a

[1, 2, 3]
[5, 2, 3]
[5, 2, 3]

可以看下不加函数就直接对列表进行操作的情况：
a = [1, 2, 3]
temp = a
temp[0] = 5
print temp
print a

[5, 2, 3]
[5, 2, 3]

再看一下对于单个数值的情况：
a = 1
temp = a
temp = 5
print temp
print a

5
1

再来改变源对象的值，看看所赋值对象的变化：
a = 1
temp = a
a = 5
print temp
print a

1
5

对于列表的情况：

a = [1, 2, 3]
temp = a
a[0] = 5
print temp
print a

[5, 2, 3]
[5, 2, 3]

对于函数的情况下：

a = [1, 2, 3]

def list_info(list_m):
temp = list_m
list_m[0] = 5
return temp
print a
print list_info(a)
print a

[1, 2, 3]
[5, 2, 3]
[5, 2, 3]