Python binding environment on locals

Question

In the following:

a = 3
def b():
    # global a; // will error without this statement
    a = a + 1
    print a

It will error unless I add a global. It seems in this sense that python evaluates the LHS of the expression first (creating a new local for a) rather than the RHS first (which equals 4) and then assigning it to the local a. In other words, like:

local a <-- local a + 1
             ^
            doesnt exist so look up in parent environment
local a <-- global a + 1
local a <--    3     + 1

I’m just curious why that approach isn’t used in python as the default. For example, in C it uses this pattern:

// file.c
#include <stdio.h>

int a=3;
int main(void)
{
    a = a + 1;
    printf("%dn", a);
}

$ gcc file.c -o file; ./file
4

Asked By: carl.hiass

||

Source

Answer 1

There is no official explanation, but I can think of two reasons

Since a = a + 1 is an assigment, it refers to the local variable a, not the global one (unless otherwise specified). Since you have not declared a local a, it is implicitly defined, but not initialized (something similar happens in javascript too, and is also a common source of confusion). In C you would not have that misunderstanding, it’s a static language, you would have defined a local int a if it existed.
In python you could have defined a function c() inside function b(), which would bind to the a variable inside b, not the global a. C doesn’t have closures, so this is not useful.

Answered By: blue_note

Python binding environment on locals

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