Scope of python variable in for loop
Solution 1:
The for loop iterates over all the numbers in range(10)
, that is, [0,1,2,3,4,5,6,7,8,9]
.
That you change the current value of i
has no effect on the next value in the range.
You can get the desired behavior with a while loop.
i = 0
while i < 10:
# do stuff and manipulate `i` as much as you like
if i==5:
i+=3
print i
# don't forget to increment `i` manually
i += 1
Solution 2:
Analogy with C code
You are imagining that your for-loop
in python is like this C code:
for (int i = 0; i < 10; i++)
if (i == 5)
i += 3;
It's more like this C code:
int r[] = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
for (int j = 0; j < sizeof(r)/sizeof(r[0]); j++) {
int i = r[j];
if (i == 5)
i += 3;
}
So modifying i
in the loop does not have the effect you expect.
Disassembly example
You can look at the disassembly of the python code to see this:
>>> from dis import dis
>>> def foo():
... for i in range (0,10):
... if i==5:
... i+=3
... print i
...
>>> dis(foo)
2 0 SETUP_LOOP 53 (to 56)
3 LOAD_GLOBAL 0 (range)
6 LOAD_CONST 1 (0)
9 LOAD_CONST 2 (10)
12 CALL_FUNCTION 2
15 GET_ITER
>> 16 FOR_ITER 36 (to 55)
19 STORE_FAST 0 (i)
3 22 LOAD_FAST 0 (i)
25 LOAD_CONST 3 (5)
28 COMPARE_OP 2 (==)
31 POP_JUMP_IF_FALSE 47
4 34 LOAD_FAST 0 (i)
37 LOAD_CONST 4 (3)
40 INPLACE_ADD
41 STORE_FAST 0 (i)
44 JUMP_FORWARD 0 (to 47)
5 >> 47 LOAD_FAST 0 (i)
50 PRINT_ITEM
51 PRINT_NEWLINE
52 JUMP_ABSOLUTE 16
>> 55 POP_BLOCK
>> 56 LOAD_CONST 0 (None)
59 RETURN_VALUE
>>>
This part creates a range between 0 and 10 and realizes it:
3 LOAD_GLOBAL 0 (range)
6 LOAD_CONST 1 (0)
9 LOAD_CONST 2 (10)
12 CALL_FUNCTION 2
At this point, the top of the stack contains the range.
This gets an iterator over the object on the top of the stack, i.e. the range:
15 GET_ITER
At this point, the top of the stack contains an iterator over the realized range.
FOR_ITER begins iterating over the loop using the iterator at the top of th estack:
>> 16 FOR_ITER 36 (to 55)
At this point, the top of the stack contains the next value of the iterator.
And here you can see that the top of the stack is popped and assigned to i
:
19 STORE_FAST 0 (i)
So i
will be overwritten regardless of what you do in the loop.
Here is an overview of stack machines if you haven't seen this before.