def s(arr):
    if len(arr) == 1:
        return arr[0]

    return arr[0]   s(arr[1:])

Above code is to find the sum of a list using recursion. So when I call the function again by giving a slicing list isn't the length of the list always gonna be same after every recursion call? Then how is my base condition is satisfying?

CodePudding user response：

Easy solution to get a sum of a list is to use sum() function, e.g. list_sum = sum([1, 2, 3])

CodePudding user response：

At every iteration, your method s calls recursively itself on a shrinked version of the arr parameter: this is done at s(arr[1:]) statement.

1. First call: arr points to [1,2,3]
2. Second call: arr points to [2,3]
3. Third call: arr points to [3]

At this point, since len([3]) == 1, the case base guard is satisfied and the return value is 3 and so on recursively, popping one activation record at time from the stack (LIFO).

A demo here: demo

CodePudding user response：

When you slice a list, it creates a new list, typically with a smaller number of elements.

You can verify it easily by using print to examine the lists:

def print_indexed_list(l):
    print(dict(enumerate(l)))

arr = ['a', 'b', 'c', 'd', 'e']

print_indexed_list(arr)
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e'}

print_indexed_list(arr[1:])
{0: 'b', 1: 'c', 2: 'd', 3: 'e'}

print_indexed_list(arr[:4])
{0: 'a', 1: 'b', 2: 'c', 3: 'd'}

print_indexed_list(arr[1:3])
{0: 'b', 1: 'c'}

print_indexed_list(arr[::2])
{0: 'a', 1: 'c', 2: 'e'}