Burhan Khalid - 2 years ago 66

Python Question

A discussion with a friend led to the following realization:

`>>> import dis`

>>> i = lambda n: n*24*60*60

>>> dis.dis(i)

1 0 LOAD_FAST 0 (n)

3 LOAD_CONST 1 (24)

6 BINARY_MULTIPLY

7 LOAD_CONST 2 (60)

10 BINARY_MULTIPLY

11 LOAD_CONST 2 (60)

14 BINARY_MULTIPLY

15 RETURN_VALUE

>>> k = lambda n: 24*60*60*n

>>> dis.dis(k)

1 0 LOAD_CONST 4 (86400)

3 LOAD_FAST 0 (n)

6 BINARY_MULTIPLY

7 RETURN_VALUE

The second example is clearly more efficient simply by reducing the number of instructions.

My question is, is there a name for this optimization, and why doesn't it happen in the first example?

Also, I'm not sure if this is a duplicate of Why doesn't GCC optimize a*a*a*a*a*a to (a*a*a)*(a*a*a)? ; if it is please explain a bit further as it applies to Python.

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Answer Source

This optimization technique is called constant folding.

The reason for constant folding occurring in the latter code but not in the former is that Python has dynamic typing, and while in mathematics a product of real numbers is *commutative* and freely *associative*, it is not so in Python in the general case, because neither do all variables contain real numbers, nor can one know the types beforehand.

Multiplication in Python is left-associative - `24 * 60 * 60 * n`

behaves like `(((24 * 60) * 60) * n)`

, which in turn implicitly executes like

```
(24).__mul__(60).__mul__(60).__mul__(n)
```

or

```
(n).__rmul_((24).__mul__(60).__mul__(60))
```

whereas `n * 24 * 60 * 60`

which is `(((n * 24) * 60) * 60)`

*can* behave like

```
n.__mul__(24).__mul__(60).__mul__(60)
```

or

```
(24).__rmul__(n).__mul__(60).__mul__(60)
```

Since we cannot know the behaviour of `n.__mul__`

beforehand, we cannot fold a constant in the latter case. Consider this example of a funny class that returns a subclass of `int`

that defines `__mul__`

/`__rmul__`

as returning the sum of the operands instead of product:

```
class MultiplyAsAdd(int):
def __mul__(self, other):
return MultiplyAsAdd(self + other)
def __rmul__(self, other):
return MultiplyAsAdd(other + self)
```

Then

```
>>> (lambda n: 24*60*60*n)(MultiplyAsAdd(5))
86405
>>> (lambda n: n*24*60*60)(MultiplyAsAdd(5))
149
```

Clearly it'd be wrong for Python to parenthesize the product as `n*(24*60*60)`

in the latter case.

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