Python3.x函数的值不正确

我不明白为什么我的代码给了我错误的值。。作为Python的初学者，我正在学习def

所以我的代码是：

def secret_formula(started):

  jelly_beans = started * 500
  jars = jelly_beans / 1000
  crates = jars / 100
  return jelly_beans, jars, crates


start_point = 10000
beans, jars, crates = secret_formula(start_point)

print(f"With a starting point of: {start_point}")
print(f"We'd have {beans} beans, {jars} jars, and {crates} crates.")

start_point = start_point / 10

print("We can have also do that this way:")
print("We'd have {} beans, {} jars, and {} crates.".format(beans, jars, crates))

当我运行我的程序时，我的答案是：

起点：10000

我们将有5000000颗豆子，5000.0罐和50.0箱

我们也可以这样做：

我们将有5000000颗豆子，5000.0罐和50.0箱

但我认为这应该是：

起点：10000

我们将有5000000颗豆子，5000.0罐和50.0箱

我们也可以这样做：

我们有50万颗豆子，500罐，5箱

像这样的事。。。因为起点=起点/10对吗

我做错了什么

Obs：是的，出于测试原因，我使用了不同的打印方法。

您需要回忆一下秘方，如：

测试代码： 结果： 你需要回忆一下秘方，比如：

测试代码： 结果：

---

更改起始点的值并不意味着重新计算豆子、板条箱和罐子的值。您必须重新分配它们以获得正确的值

---

更改起始点的值后，尝试重复调用秘密公式的行。

更改起始点的值并不意味着重新计算豆子、板条箱和罐子的值。您必须重新分配它们以获得正确的值

---

更改起始点的值后，请尝试重复调用秘密公式的行。

在您第一次运行秘密公式时，豆子、罐子和板条箱的值已经指定。更改start_point将更新变量start_point的值，而不是其他变量的值

通过再次运行secret_公式，您将获得所需的新值

def secret_formula(started):
    jelly_beans = started * 500
    jars = jelly_beans / 1000
    crates = jars / 100
    return jelly_beans, jars, crates


start_point = 10000
beans, jars, crates = secret_formula(start_point)

print(f"With a starting point of: {start_point}")
print(f"We'd have {beans} beans, {jars} jars, and {crates} crates.")

start_point = start_point / 10
beans, jars, crates = secret_formula(start_point)

print("We can have also do that this way:")
print("We'd have {} beans, {} jars, and {} crates.".format(beans, jars, crates))

---

豆子、罐子、板条箱的价值在您第一次运行secret_公式时已指定。更改start_point将更新变量start_point的值，而不是其他变量的值

通过再次运行secret_公式，您将获得所需的新值

def secret_formula(started):
    jelly_beans = started * 500
    jars = jelly_beans / 1000
    crates = jars / 100
    return jelly_beans, jars, crates


start_point = 10000
beans, jars, crates = secret_formula(start_point)

print(f"With a starting point of: {start_point}")
print(f"We'd have {beans} beans, {jars} jars, and {crates} crates.")

start_point = start_point / 10
beans, jars, crates = secret_formula(start_point)

print("We can have also do that this way:")
print("We'd have {} beans, {} jars, and {} crates.".format(beans, jars, crates))

---

您忘记第二次调用secret\u公式更改start\u point的值不会影响其他变量！我想他没有忘记。我认为他更倾向于假设更改该公式的输入将自动更新结果，这显然没有发生。您忘记再次调用secret_公式更改start_point的值不会影响其他变量！我想他没有忘记。我认为他更倾向于假设将输入更改为该公式将自动更新结果，这显然不是那么简单！谢谢大家!！太简单了！谢谢大家!！

With a starting point of: 10000
We'd have 5000000 beans, 5000.0 jars, and 50.0 crates.
We can have also do that this way:
We'd have 500000.0 beans, 500.0 jars, and 5.0 crates.

def secret_formula(started):
    jelly_beans = started * 500
    jars = jelly_beans / 1000
    crates = jars / 100
    return jelly_beans, jars, crates


start_point = 10000
beans, jars, crates = secret_formula(start_point)

print(f"With a starting point of: {start_point}")
print(f"We'd have {beans} beans, {jars} jars, and {crates} crates.")

start_point = start_point / 10
beans, jars, crates = secret_formula(start_point)

print("We can have also do that this way:")
print("We'd have {} beans, {} jars, and {} crates.".format(beans, jars, crates))