Behaviour of increment and decrement operators in Python

Question

I notice that a pre-increment/decrement operator can be applied on a variable (like ++count). It compiles, but it does not actually change the value of the vari

Answer 1

Python does not have these operators, but if you really need them you can write a function having the same functionality.

def PreIncrement(name, local={}):
    #Equivalent to ++name
    if name in local:
        local[name]+=1
        return local[name]
    globals()[name]+=1
    return globals()[name]

def PostIncrement(name, local={}):
    #Equivalent to name++
    if name in local:
        local[name]+=1
        return local[name]-1
    globals()[name]+=1
    return globals()[name]-1

Usage:

x = 1
y = PreIncrement('x') #y and x are both 2
a = 1
b = PostIncrement('a') #b is 1 and a is 2

Inside a function you have to add locals() as a second argument if you want to change local variable, otherwise it will try to change global.

x = 1
def test():
    x = 10
    y = PreIncrement('x') #y will be 2, local x will be still 10 and global x will be changed to 2
    z = PreIncrement('x', locals()) #z will be 11, local x will be 11 and global x will be unaltered
test()

Also with these functions you can do:

x = 1
print(PreIncrement('x'))   #print(x+=1) is illegal!

But in my opinion following approach is much clearer:

x = 1
x+=1
print(x)

Decrement operators:

def PreDecrement(name, local={}):
    #Equivalent to --name
    if name in local:
        local[name]-=1
        return local[name]
    globals()[name]-=1
    return globals()[name]

def PostDecrement(name, local={}):
    #Equivalent to name--
    if name in local:
        local[name]-=1
        return local[name]+1
    globals()[name]-=1
    return globals()[name]+1

I used these functions in my module translating javascript to python.