Why Can't `exec` Directly Update Local Variables in Python 3?

How Local Variables Can't Be Updated Directly with exec

The exec call in Python is a powerful tool for executing code dynamically, but it comes with limitations regarding the modification of local variables.

Consider the following code:

def f():
    a = 1
    exec("a = 3")
    print(a)

f()

One might expect this code to print 3, but it actually prints 1. This is because in Python 3, local variables are not stored in a dictionary, but an array with indices determined at compile time. The exec function can't safely modify local variables without interfering with this optimization.

Solution: Using Local Dictionary with exec

To modify local variables using exec, you need to explicitly pass a local dictionary. For example:

def foo():
    ldict = {}
    exec("a = 3", globals(), ldict)
    a = ldict['a']
    print(a)

This executes the code within the local dictionary (ldict), which is distinct from the function's local variable array. The modified variable can then be returned to the function's scope by accessing the local dictionary.

Python 2 Behavior

In Python 2, exec could modify local variables without passing an explicit dictionary because it treated namespaces that used exec without globals/locals arguments as "unoptimized." However, this is not the case in Python 3.

Therefore, it's important to remember that when using exec, local variables can only be updated by creating and passing a local dictionary to avoid any potential conflicts with the compiler's optimizations.

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