指数函数返回问题-python

我正在研究这个问题-

 modify powerOfTwo() to meet the conditions below
#    - accept 1 integer parameter; an exponent
#    - print to the screen 2 raised to the exponent argument, example:
#    --- "2 to the power of 2 is 4"
#    - also return the result (for the example, you would return 4)

我的代码如下：

def powerOfTwo(exp):
    x = int(input("please enter a number for what power of 2: "))
    e = (2**x)
    print("2 to the power of",x, "is: ",e,)



    return (2**x)

当运行程序时，我教授的代码检查我的代码的准确性，我相信它检查得不正确。程序返回以下内容，其中包括我自己输入的“5”：

---

教授的程序会生成一个随机数，输入到函数中。所以你不能用“输入”来输入你自己的号码。您必须使用值“exp”。

教授的代码生成一个随机数并将其放入函数中。为了检查函数是否正确运行，只需输入生成的随机数

此外，您似乎还没有在您提供的内容中初始化分数：

q2s = 0
e = random.randint(3,11)
score=0

---

我明白你的意思。但是，输入函数允许我接收用户输入…而“x=exp”（“请输入一个2的幂：”）不允许。您不允许进行用户输入。您教授的程序通过exp参数指定输入

q2s = 0
e = random.randint(3,11)
try:
    print("\nq2-1:  Checking powerOfTwo(" + str(e) + ")")
    p2 = powerOfTwo(e)
    print("+3 function call w/ correct # of params is present")
    score += 3
    q2s += 3
    if isinstance(p2, int):
        print("+2 return value is correct type:",type(p2))
        score += 2
        q2s += 2
        if p2 == (2 ** e):
            print("+2 return value is correct:",p2)
            score += 2
            q2s += 2
        else:
            print("-- return value is incorrect")
            print("Expected: ",2 ** e)
            print("Returned: ",p2)

        e = random.randint(12,22)
        print("\nq2-2:  Checking powerOfTwo(" + str(e) + ")")
        p2 = powerOfTwo(e)
        if p2 == (2**e):
            print("+2 return value is correct:",p2)
            score += 2
            q2s += 2
        else:
            print("-- return value is incorrect")
            print("Expected: ",2**e)
            print("Returned: ",p2)
    else:
        print("-- return value is incorrect type:",type(p2))
except:
    print("**Something is wrong with powerOfTwo()")
    printErr()

q2s = 0
e = random.randint(3,11)
score=0