Question

我创建了一个函数，我想在列表中附加一个数字：

variableToChange[0] = true;

为什么我必须返回一个功能？它可以使用和不使用def num(f): list1.append(i) return list1 list1 = [] i = 1 print "Now list1 is %s and i is %d" % (list1, i) num(list1) i += 1 print "Now list1 is %s and i is %d" % (list1, i) num(list1) i += 1 print "Now list1 is %s and i is %d" % (list1, i) print list1 print i。

如果没有达到return语句，我被告知该函数返回return。但是，即使我没有输入这个返回语句，上面提到的函数仍然有效。

Answer 1

在提供的示例中，返回值是不必要的，因为append列表方法以及您的函数也可以通过副作用进行操作。仅为副作用调用的函数不是数学意义上的函数（域值到codomain值的映射），它们用于更改可变对象的状态，在这种情况下，通过将项附加到列表。虽然这些函数可以返回一个值，但是Python有一个约定，该函数纯粹为了副作用调用，例如list.append，只返回None，表示没有有用的返回值。 / p>

如果你的函数没有副作用，你需要有一个return语句才能使它有用。例如，比较：

def add(a, b):
    return a + b

......在语法上同样合法，但很无用：

def add(a, b):
    a + b
    # without a return statement, None is returned,
    # and the calculated sum discarded

Answer 2

我发现你不了解函数是如何工作的，所以我在你的代码中添加了一些注释来解释一下，但我建议你进一步阅读Python tutorial about functions和wiki article以获得理解。

另外，我省略了许多细节，不会使解释过载。重要的是有不可变的（即你的例子中的整数，i）和Python中的可变（即你的例子中的列表，list1类型，并且根据这种情况，行为会有所不同。

def num(f):
    #Here the argument you pass to the function is named 'f'
    #and you don't use it

    #The next line uses 'list1', that is defined in global scope
    #since you didn't redefined this name inside the function
    #Variable 'i' is also the one in global scope for same reasons
    list1.append(i)

    #Here you return 'list1', though you don't use this value
    #further in your program. Indeed, you would not write a return
    #statement the function would return 'None' as the return value
    return list1

#Here you define 'list1' in global scope, and it will be used 
#inside 'num' function, even without providing it as the argument
list1 = []

#Here you define 'i' in global scope, and it will be used 
#inside 'num' function
i = 1

#Here you print 'i' and 'list' from global scope
print "Now list1 is %s and i is %d" % (list1, i)   

#Here you call 'num' function and 'list1' provided as argument
#is assigned to 'f' inside the function, but you didn't used it and
#and instead used names from global scope - that's why it works in
#this way (however it is wrong use of function)
#With 'list1.append(i)' the 'list1' is modified in place so it
#doesn't matter if it is returned or not
num(list1)

#As to 'num' return value, it will be the same value as 'list1', but
#you don't use it here, to use it it needs to be assigned with '=' to
#some variable, i.e. 'list2=num(list1)', though in fact 'list1' and 'list2'
#will be the same all the time due to Python internals, but let's skip the
#details of this.

#You can see that the value returned is not 'None' - add the 
#following line here:
print(num(list1))
#and run your program, the output will show you that it's a list returned.
#then remove the 'return' line in your function and run program again
#the output here will show, that is's 'None' that was returned.

所以要解决函数中明显的错误：

def num(f):
    f.append(i)
    return f

但是我仍然在全局范围内使用而不是作为参数传递，所以更好：

 def num(f_var,i_var):
    f_var.append(i_var)
    return f_var

虽然该列表将在适当的位置进行修改，但您并不需要将其返回在你的特定例子中，所以：

 def num(f_var,i_var):
    f_var.append(i_var)

 list1=[]
 i=1

 num(list1,i)

也会奏效。

Answer 3

你没有使用返回值，所以它没有区别，你返回的是什么。你也没有使用这个论点。你可能想写

def append_num(f, i):
    f.append(i)

并使用两个参数：

append_num(list1, i)

为什么我必须返回＆＃39;一个功能？

3 个答案: