I have this code, which prints 1:
s = 0
def dfs(n):
global s
if n > 10:
return 0
s += dfs(n + 1)
return n
dfs(0)
print(s)
If I modify dfs like this：
def dfs(n):
global s
if n > 10:
return 0
i = dfs(n + 1)
s += i
return n
it will print 55
I know what is a better way to write the dfs. I just want to know why the value of s is different after the call of two dfs
Python is interpreted and executed from top to bottom, so in the first version you have:
s += dfs(n + 1)
which is exact as:
s = s + dfs(n + 1)
So when you do this recursively you have on stack these commands:
s = 0 + dfs(1) # <-- dfs(1) will return 1
s = 0 + dfs(2) # <-- dfs(2) will return 2
...
s = 0 + dfs(9) # <-- dfs(9) will return 9
s = 0 + dfs(10) # <-- dfs(10) will return 10
s = 0 + dfs(11) # <-- dfs(11) will return 0
So when you observe, the last step is s = 0 + 1 and you see the 1 as final result.
The second version
i = dfs(n + 1)
s += i
You assign to s after dfs(n + 1) is evaluated so you see the final answer 55
NOTE: If you rewrite the first version s += dfs(n + 1) to s = dfs(n + 1) + s you will see the result 55 too.