def add(a):
a = a + 1
print(a)
if a < 10:
add(a)
print(a)
add(1)
I don't understand why the value of a started reducing after 10 when it should've stopped at 10 even without an else statement. Can anyone explain the reason?
def add(a):
a = a + 1
print(f"#1, a: {a}, id: {id(a)}")
if a < 10:
add(a)
print(f"#2, a: {a}, id: {id(a)}")
add(1)
# stdout:
#1, a: 2, id: 9801280
#1, a: 3, id: 9801312
#1, a: 4, id: 9801344
#1, a: 5, id: 9801376
#1, a: 6, id: 9801408
#1, a: 7, id: 9801440
#1, a: 8, id: 9801472
#1, a: 9, id: 9801504
#1, a: 10, id: 9801536
#2, a: 10, id: 9801536
#2, a: 9, id: 9801504
#2, a: 8, id: 9801472
#2, a: 7, id: 9801440
#2, a: 6, id: 9801408
#2, a: 5, id: 9801376
#2, a: 4, id: 9801344
#2, a: 3, id: 9801312
#2, a: 2, id: 9801280
And if you disassemble:
import dis
dis.dis(add)
The bytecode doesn't show any BINARY_SUBTRACT operation.
The if statement recursively executes add() until the last line of the function, which then "waits" for the older add() calls to end.
Each n call was called by the n-1 one, as it is shown on the stdout: this results in all these waiting stacked calls being executed in a LIFO order.
At the end, a's value is not being decremented, instead is printed with the old values from each previous stacked call, in a "chiastic" fashion.
This kind of recursion can be used for printing pyramids (double top) on the stdout:
def print_pyramid(a):
a = a + 1
print(a*'*')
if a < 5:
print_pyramid(a)
print(a*'*')
print_pyramid(0)
# stdout:
*
**
***
****
*****
*****
****
***
**
*
print(1);print("bob");print(1);prints 1 twice?aas it steps out of the recursion. Recursion stops when you reacha = 10because of theif a < 10:, then your second print runs for the current iteration, then it can run for the previous iteration and so on.print(a)twice, so you will see every value ofaprinted twice.