Function parameter not writing to variable

[Karp]

Hi,

I just started learning python and have run into a problem with writing a function recombining a list of separated characters into a string. When running the function nothing gets added to the variable that is put in as the combined parameter (output_string), resulting in nothing getting printed in print(output_string). Why no work?

input_list = ["h", "e", "l", "l", "o"]
output_string = ""

def list_comb(separated, combined):
    for letter in separated:
        combined = combined + letter


list_comb(input_list, output_string)

print(output_string)

Thanks!

[menator01]

Your not returning or printing anything from the function call

input_list = ["h", "e", "l", "l", "o"]

def list_comb(mylist):
    print(f'Print in the function -> {"".join(mylist)}')
    return ''.join(mylist)

print(f'Returned from function call -> {list_comb(input_list)}')

[snippsat]

To fix you original code.

def list_comb(separated, combined):
    for letter in separated:
        combined += letter
    return combined

input_list = ["h", "e", "l", "l", "o"]
output_string = ""
result = list_comb(input_list, output_string)
print(result)

Output:
hello

As posted over this is what "".join(lst) do,as training it's a ok task.

[Karp]

Your not returning or printing anything from the function call

input_list = ["h", "e", "l", "l", "o"]

def list_comb(mylist):
    print(f'Print in the function -> {"".join(mylist)}')
    return ''.join(mylist)

print(f'Returned from function call -> {list_comb(input_list)}')

Thanks! Makes sense

Out of curiosity though, the bellow function works even though there is not a print or return (to me it feels like it is the same as the previous example where it changes external data outside of the function when running it. I assume they are different, but feel like they are doing the same thing (only difference being one changing a string and one changing a list)). What is the difference that makes one work and not the other?

input_string = "hello"
output_list = []

def separate_char(combined, separated):
    for symbol in combined:
        separated.append(symbol)

separate_char(input_string, output_list)
print(output_list)

Output:
['h', 'e', 'l', 'l', 'o']

Thanks again! Hope the question makes sense

[deanhystad]

Lists are mutable (they can be modified) and Python passes arguments by object. When you pass a mutable object to a function, the function can modify the mutable object. Since the changes are to the object, those changes are seen outside the function. In your example, you passed a mutable object, "output_list" to a function. The function modified "output_list", adding strings to the list. Now anyone looking at the list will see the modified contents.

Strings are not mutable. You cannot change a string. When you add two strings it creates a new string. In your first example, "combined = combined + letter" does not modify the string referenced by "combined", it creates a brand new string and assigns "combined" to reference the new string. Since the original blank string passed to the function cannot be changed, and output_string has not been reassigned to reference a different object, output_string still reference the original, blank string.

[Karp]

Lists are mutable (they can be modified) and Python passes arguments by object. When you pass a mutable object to a function, the function can modify the mutable object. Since the changes are to the object, those changes are seen outside the function. In your example, you passed a mutable object, "output_list" to a function. The function modified "output_list", adding strings to the list. Now anyone looking at the list will see the modified contents.

Strings are not mutable. You cannot change a string. When you add two strings it creates a new string. In your first example, "combined = combined + letter" does not modify the string referenced by "combined", it creates a brand new string and assigns "combined" to reference the new string. Since the original blank string passed to the function cannot be changed, and output_string has not been reassigned to reference a different object, output_string still reference the original, blank string.

Thank you so much! Great explanation