python：如何更改函数输入参数的值？

Question

I tried to modify the value of a string inside a function, like below:

>>> def appendFlag(target, value):
...     target += value
...     target += " "
...
>>> appendFlag
<function appendFlag at 0x102933398>
>>> appendFlag(m,"ok")
>>> m
''

Well, seems the "target" is only changed within the function, but how to make the new value viable outside the function? Thanks.

Answer 1

This is handled in python by returning.

def appendFlag(target, value):
   target += value
   target += " "
   return target

you can use it like this:

m = appendFlag(m,"ok")

you can even return several variables like this:

def f(a,b):
   a += 1
   b += 1
   return a,b

and use it like this:

a,b = f(4,5)

Answer 2

You need to use an object that can be modified

>>> m = []
>>> def appendFlag(target, value):
...     target.append(value)
...     target.append(" ")
...
>>> appendFlag(m, "ok")
>>> m
['ok', ' ']

Answer 3

I tried to modify the value of a string inside a function, like below:

>>> def appendFlag(target, value):
...     target += value
...     target += " "
...
>>> appendFlag
<function appendFlag at 0x102933398>
>>> appendFlag(m,"ok")
>>> m
''

Well, seems the "target" is only changed within the function, but how to make the new value viable outside the function? Thanks.

Answer 4

The variables inside the function has got only function scope, if you change the value of the variable from inside the function it wont get reflected outside the function. If you want to change the value of the global variable m, update it from outside the function as follows

 def appendFlag(target, value):
        target += value
        target += " "
        return target
  m=''
  m=appendFlag(m,'ok')

Answer 5

Mutable data types, such as lists or dictionaries, can be changed from within the function without returning:

def duplicate_last(l):
    l.append(l[-1])

l = [1, 2, 3]
duplicate_last(l)

The reason for that is different memory management for mutable and immutable objects in python.

Answer 6

Your function needs to return the new value. The target variable only has scope within the appendFlag function.

def appendFlag(target, value):
   ...
   return target

m = appendFlag(m, "ok")

Answer 7

Make a new class and set the target as the member of class. Then try setattr(class, 'target', value).