Why local variables referenced before assignment?-CodePudding

I made the two functions below in Python 3. The first function test_list works fine with the list a without error. I can modify the list element in that function.

However, the second funciton test_int will pop an error local variable 'b' referenced before assignment. Why can't I do this to the variable b?

a=[1,2,3]
def test_list():
    a[0]=2
    return a

b = 2
def test_int():
    b  = 1
    return b

CodePudding user response：

b = 1 is equivalent to b = b.__iadd__(1); since the name b gets bound by that statement (l.h.s.), the compiler infers that b is a local variable; however, the r.h.s. expression contains the name b as well and thus you get the error "local variable 'b' referenced before assignment" (the local name b doesn't refer to anything at the time when the expression is evaluated).

a[0] = 2 on the other hand is equivalent to a.__setitem__(0, 2) and thus involves no name binding.

If you wish to modify a global name binding within a function, you can use the keyword global; there's also the equivalent nonlocal for names contained in an outer, but not the global, scope.

CodePudding user response：

b is globally scoped. You are attempting to modify it in a local context, ie in test_int. You need the global keyword if you wish to modify a globally scoped variable in a local context. Read about variable scoping in python here.

The reason the first function test_list works is because you are not modifying the global variable itself, but only the contents. Here's a good explanation of what happens when you modify global variables: https://stackoverflow.com/a/31437415/14715054

This will fix your problem:

a=[1,2,3]
def test_list():
    a[0]=2
    return a

b = 2
def test_int():
    global b
    b  = 1
    return b