为什么变量在 Python 中按值传递时具有相同的 id

Question

I have been studying functions in python and how the variables are passed by values and not reference since its a lot safer. if that is the way it is then why do the variable passed around have the same ID.I initially used x=3 but then I read that how python caches variables from -5 to 256 so I used 500 but it still shows the same id. If they have the same doesn't it mean that it is the same object passed around?

---

x = 500
print(id(x))
def cat(a):
    print(id(a))

def main():
    a = x
    print(id(a))
    cat(a)
if __name__ == '__main__': main()

---

Output:

 4511355632 4511355632 4511355632

Answer 1

I have been studying functions in python and how the variables are passed by values and not reference since its a lot safer

I don't know where you read this, but that's just complete bullshit.

if that is the way it is

It isn't.

then why do the variable passed around have the same ID.

Because they point to the same object, obviously.

I initially used x=3 but then I read that how python caches variables from -5 to 256

That's a CPython implementation detail, and the exact values depend on the CPython version etc. But anyway, you can test this with any number you want, and actually just any type, the result will still be the same.

so I used 500 but it still shows the same id. If they have the same doesn't it mean that it is the same object passed around?

id(obj) , by definition, returns the object's unique identifier (actually the memory address in CPython but that's also an implementation detail), so by definition, if two objects have the same id, then they are indeed the very same object.

NB : "unique" meaning that for the current process, no other object will have the same id at the same time - once an object is garbage-collected, it's id can be reused.

FWIW, using a mutable object, it's quite easy to find out that it's not passed "by value":

def foo(lst):
    lst.append(42)

answers = []
for i in range(10):
    print("{} - before: {}".format(i, answers))
    foo(answers)
    print("{} - after: {}".format(i, answers))

0 - before: []
0 - after: [42]
1 - before: [42]
1 - after: [42, 42]
2 - before: [42, 42]
2 - after: [42, 42, 42]
3 - before: [42, 42, 42]
3 - after: [42, 42, 42, 42]
4 - before: [42, 42, 42, 42]
4 - after: [42, 42, 42, 42, 42]
5 - before: [42, 42, 42, 42, 42]
5 - after: [42, 42, 42, 42, 42, 42]
6 - before: [42, 42, 42, 42, 42, 42]
6 - after: [42, 42, 42, 42, 42, 42, 42]
7 - before: [42, 42, 42, 42, 42, 42, 42]
7 - after: [42, 42, 42, 42, 42, 42, 42, 42]
8 - before: [42, 42, 42, 42, 42, 42, 42, 42]
8 - after: [42, 42, 42, 42, 42, 42, 42, 42, 42]
9 - before: [42, 42, 42, 42, 42, 42, 42, 42, 42]
9 - after: [42, 42, 42, 42, 42, 42, 42, 42, 42, 42]