为什么 python 不打印我写入的所有内容？

Question

I was doing some exercises in python python and I bumped into a problem. I already printed a list reversed and I wanted to print the reversed elements in separate lines too but for some reason it only prints the first one and if I put the first print into comment it prints out the loop just fine. Why?

def main():
    pass
if __name__ == '__main__':
    main()

list1 = [10, 20, 30, 40, 50]

def reverse_list(l1):
    l2 = []
    for i in range(len(l1)):
        l2.append(l1.pop())
    return l2

print(reverse_list(list1))

pass

def reverse_print(l):
    listr = reversed(l)
    for j in listr:
        print(j)

reverse_print(list1)

print("Complete...")

Answer 1

In python, data types are divided into mutable and immutable types:

• Mutable object : the value in the memory pointed to by the object can be changed
• Immutable object: The value in the memory pointed to by the object cannot be changed, so when the value pointed to by the variable changes, it is equivalent to copying the original value and storing it in a new address, and the variable points to this new address.

a = [1,2,3,4]
a_ref = a
print(a) # [1,2,3,4]
a_ref[0] = 5
print(a) # [5,2,3,4]

It can be seen from the above code

1. a, a_ref are exactly the same
2. a and a_ref have the same memory location
3. modification to a_ref, a will also be changed

for example:

a = [1,2,3,4]
def somefunc(a2):
    a2.append(5)
print(a) # [1,2,3,4]
somefunc(a)
print(a) # [1,2,3,4,5]

Normally, it won't change a by calling somefunc(a)
BUT
Since a2 is a reference to a and points to the same memory location, when a2 is modified, it will also be modified to a

in your code:
when you calling reverse_list(list1) , l1 is a reference to list1, when you pop the element from l1, it will also pop the element in list1

so after called reverse_list(list1) , list1 is empty, that why reverse_print(list1) do not print anything

Sollution

def reverse_list(l1):
    l1_copy = l1[:] # copy the whole list
    l2 = []
    for i in range(len(l1_copy)):
        l2.append(l1_copy.pop())
    return l2

when you call l1_copy = l1[:] than l1_copy won't point to same memory location so l1 and list1 won't modified when modifica l1_copy

PS: english is my second langage, so there maybe have some gramma mistake, please excuse me

Answer 2

at the second print list1 has been emptied out from using the pop() method in reverse_list .

to reverse a list perhaps you should use the reverse() method like the following example:

List1 = [1,2,3]
List1.reverse()
print(List1) # [3,2,1]