python - 如何使用整数变量作为函数的参数？

Question

我正在尝试编写一段代码来计算三个不同候选人的选票，并且我正在使用一个函数，该函数使用变量名称(A、B 或 C)作为参数。

我试图让它在计算该候选人的选票时，它会调用该函数将该候选人的变量增加 1。但是，无论我尝试过何种方式，所有 3 名候选人都将始终计入 0 票，除非我完全删除该功能。

我尝试了几种不同的方法来使变量成为全局变量，但它们都给出了相同的结果。

A = 0
B = 0
C = 0

def after_vote(letter):
    letter = letter + 1
    print("Thank you for your vote.")

def winner(winner_letter, winner_votes):
    print("The winner was", winner_letter, "with", str(winner_votes), "votes.")

while True:
    vote = input("Please vote for either Candidate A, B or C. ").upper()
    if vote == "A":
        after_vote(A)
    elif vote == "B":
        after_vote(B)
    elif vote == "C":
        after_vote(C)
    elif vote == "END":
        print("Cadidate A got", str(A), "votes, Candidate B got", str(B), "votes, and Candidate C got", str(C), "votes.")
        if A > B and A > C:
            winner("A", A)
            break
        elif B > A and B > C:
            winner("B", B)
            break
        elif C > A and C > B:
            winner("C", C)
            break
        else:
            print("There was no clear winner.")
            break
    else:
        print("Please input a valid option.")

Answer 1

首先，这个想法是错误的。您不想处理全局变量并传递名称。可以这样做，但这是个坏主意。

更好的选择是将要修改的变量传递给函数。但是，整数的诀窍在于它们是不可变的，因此您不能像在 C 中那样传递要由函数修改的整数。

剩下的是:

• 传递一个值给函数，从函数返回修改后的值；或
• 将一个包含值的可变对象传递给函数

这就是理论，下面是如何去做...

方案一:传一个值，返回修改后的值

def after_vote(value):
    print("Thank you for your vote.")
    # Instead of modifying value (impossible), return a different value
    return value + 1

A = after_vote(A)

解决方案 2:传递一个“可变整数”

class MutableInteger:
    def __init__(value):
        self.value = value

A = MutableInteger(0)

def after_vote(count):
    # Integers cant be modified, but a _different_ integer can be put
    # into the "value" attribute of the mutable count object
    count.value += 1
    print("Thank you for your vote.")

after_vote(A)

解决方案 3:传递所有投票的(可变!)字典

votes = {'A': 0, 'B': 0, 'C': 0}

def after_vote(votes, choice):
    # Dictionaries are mutable, so you can update their values
    votes[choice] += 1
    print("Thank you for your vote.")

after_vote(votes, "A")

解决方案 4(最糟糕的!):实际执行您要求的

def after_vote(letter):
    # Global variables are kept in a dictionary. globals() returns that dict
    # WARNING: I've never seen code which does this for a *good* reason
    globals()[letter] += 1
    print("Thank you for your vote.")