Question

所以我已经在Python 3.7中编写了一个简单的计算器，但是它不适用于我的函数。变量“输出”显示为灰色，但我不知道为什么。

n1 = 0
n2 = 0
op = 0
res = 0
output = 0

# DEFINING FUNCTIONS

def askAndStore(aasText,var2save):
    print(aasText)
    var2save = input()

def calculate(num1,num2,op,output):
    if op == "add":
        output = float(num1) + float(num2)
    elif op == "sub":
        output = float(num1) - float(num2)
    elif op == "mul":
        output = float(num1) * float(num2)
    elif op == "div":
        output = float(num1) / float(num2)
    else:
        print("Error: Unknown operation")


# PROGRAM MAIN

askAndStore("What is the first number?", n1)
askAndStore("What is the second number?", n2)
askAndStore("What is the operation?", op)
calculate(n1, n2, op, res)
print(res)

我的输出是：

What is the first number?
10
What is the second number?
5
What is the operation?
add
Error: Unknown operation
0

Process finished with exit code 0

即使我输入“ add”作为操作，它始终显示“错误：未知操作”。有什么想法为什么不起作用？

Answer 1

n1 = float(input("What is the first number? "))
n2 = float(input("What is the second number? "))
op = str(input("What is the operation? "))

# DEFINING FUNCTIONS


def calculate(num1, num2, opr):
    if opr == "add":
        output = num1 + num2
    elif opr == "sub":
        output = num1 - num2
    elif opr == "mul":
        output = num1 * num2
    elif opr == "div":
        output = num1 / num2
    else:
        print("Error: Unknown operation")
    return output

print(calculate(n1, n2, op))

Answer 2

var2save对于askAndStore函数而言是本地的。要从函数外部访问它，必须return。稍作修改：

def calculate(num1,num2,op):
    if op == "add":
        output = float(num1) + float(num2)
    elif op == "sub":
        output = float(num1) - float(num2)
    elif op == "mul":
        output = float(num1) * float(num2)
    elif op == "div":
        output = float(num1) / float(num2)
    else:
        raise ValueError("Unknown operation " + op) 
    return output

n1 = input("What is the first number?")
n2 = input("What is the second number?")
op = input("What is the operation?")
output = calculate(n1, n2, op)
print(output)

Answer 3

我猜您是一个C程序吗？或类似的东西，在python中，您不能将过去的值重新分配给这样的函数，当您将参数传递给函数时，它只是指向此PyObject的指针，并且您不能更改此指针地址，只能是值（重新分配变量对此将永远无效）

如您所见，您可以在这些情况下使用return语句

def calculate(num1,num2,op):
    output = None
    if op == "add":
        output = float(num1) + float(num2)
    elif op == "sub":
        output = float(num1) - float(num2)
    elif op == "mul":
        output = float(num1) * float(num2)
    elif op == "div":
        output = float(num1) / float(num2)
    else:
        print("Error: Unknown operation")
    return output

# PROGRAM MAIN

n1 = input("What is the first number?")
n2 = input("What is the second number?")
op = input("What is the operation?")
res = calculate(n1, n2, op)
print(res)

看看复杂的对象确实可以这样工作，例如list

def f(l):
    l.append(1)
l = []
f(l)
print(l) # [1]

但是您不能这样重新分配指针：

def f(l):
    l = [1]
l = []
f(l)
print(l) # []

基本计算器无法使用我的功能

3 个答案: