Behaviour of increment and decrement operators in Python

Question

How do I use pre-increment/decrement operators (++, --), just like in C++?

Why does ++count run, but not change the value of the variable?

Answer 1

While the others answers are correct in so far as they show what a mere + usually does (namely, leave the number as it is, if it is one), they are incomplete in so far as they don't explain what happens.

To be exact, +x evaluates to x.__pos__() and ++x to x.__pos__().__pos__().

I could imagine a VERY weird class structure (Children, don't do this at home!) like this:

class ValueKeeper(object):
    def __init__(self, value): self.value = value
    def __str__(self): return str(self.value)

class A(ValueKeeper):
    def __pos__(self):
        print('called A.__pos__')
        return B(self.value - 3)

class B(ValueKeeper):
    def __pos__(self):
        print('called B.__pos__')
        return A(self.value + 19)

x = A(430)
print(x, type(x))
print(+x, type(+x))
print(++x, type(++x))
print(+++x, type(+++x))

Answer 2

Extending Henry's answer, I experimentally implemented a syntax sugar library realizing a++: hdytto.

The usage is simple. After installing from PyPI, place sitecustomize.py:

from hdytto import register_hdytto
register_hdytto()

in your project directory. Then, make main.py:

# coding: hdytto

a = 5
print(a++)
print(++a)
b = 10 - --a
print(b--)

and run it by PYTHONPATH=. python main.py. The output will be

5
7
4

hdytto replaces a++ as ((a:=a+1)-1) when decoding the script file, so it works.

Answer 3

Python does not have pre and post increment operators.

In Python, integers are immutable. That is you can't change them. This is because the integer objects can be used under several names. Try this:

>>> b = 5
>>> a = 5
>>> id(a)
162334512
>>> id(b)
162334512
>>> a is b
True

a and b above are actually the same object. If you incremented a, you would also increment b. That's not what you want. So you have to reassign. Like this:

b = b + 1

Many C programmers who used python wanted an increment operator, but that operator would look like it incremented the object, while it actually reassigns it. Therefore the -= and += operators where added, to be shorter than the b = b + 1, while being clearer and more flexible than b++, so most people will increment with:

b += 1

Which will reassign b to b+1. That is not an increment operator, because it does not increment b, it reassigns it.

In short: Python behaves differently here, because it is not C, and is not a low level wrapper around machine code, but a high-level dynamic language, where increments don't make sense, and also are not as necessary as in C, where you use them every time you have a loop, for example.

Answer 4

A straight forward workaround

c = 0
c = (lambda c_plusplus: plusplus+1)(c)
print(c)
1

No more typing

 c = c + 1

Also, you could just write c++ and finish all your code and then do search/replace for "c++", replace with "c=c+1". Just make sure regular expression search is off.

Answer 5

In python 3.8+ you can do :

(a:=a+1) #same as ++a (increment, then return new value)
(a:=a+1)-1 #same as a++ (return the incremented value -1) (useless)

You can do a lot of thinks with this.

>>> a = 0
>>> while (a:=a+1) < 5:
    print(a)

    
1
2
3
4

Or if you want write somthing with more sophisticated syntaxe (the goal is not optimization):

>>> del a
>>> while (a := (a if 'a' in locals() else 0) + 1) < 5:
    print(a)

    
1
2
3
4

It will return 0 even if 'a' doesn't exist without errors, and then will set it to 1

Answer 6

++ is not an operator. It is two + operators. The + operator is the identity operator, which does nothing. (Clarification: the + and - unary operators only work on numbers, but I presume that you wouldn't expect a hypothetical ++ operator to work on strings.)

++count

Parses as

+(+count)

Which translates to

count

You have to use the slightly longer += operator to do what you want to do:

count += 1

I suspect the ++ and -- operators were left out for consistency and simplicity. I don't know the exact argument Guido van Rossum gave for the decision, but I can imagine a few arguments:

• Simpler parsing. Technically, parsing ++count is ambiguous, as it could be +, +, count (two unary + operators) just as easily as it could be ++, count (one unary ++ operator). It's not a significant syntactic ambiguity, but it does exist.
• Simpler language. ++ is nothing more than a synonym for += 1. It was a shorthand invented because C compilers were stupid and didn't know how to optimize a += 1 into the inc instruction most computers have. In this day of optimizing compilers and bytecode interpreted languages, adding operators to a language to allow programmers to optimize their code is usually frowned upon, especially in a language like Python that is designed to be consistent and readable.
• Confusing side-effects. One common newbie error in languages with ++ operators is mixing up the differences (both in precedence and in return value) between the pre- and post-increment/decrement operators, and Python likes to eliminate language "gotcha"-s. The precedence issues of pre-/post-increment in C are pretty hairy, and incredibly easy to mess up.

Answer 7

There are no post/pre increment/decrement operators in python like in languages like C.

We can see ++ or -- as multiple signs getting multiplied, like we do in maths (-1) * (-1) = (+1).

E.g.

---count

Parses as

-(-(-count)))

Which translates to

-(+count)

Because, multiplication of - sign with - sign is +

And finally,

-count

Answer 8

TL;DR

Python does not have unary increment/decrement operators (--/++). Instead, to increment a value, use

a += 1

More detail and gotchas

But be careful here. If you're coming from C, even this is different in python. Python doesn't have "variables" in the sense that C does, instead python uses names and objects, and in python ints are immutable.

so lets say you do

a = 1

What this means in python is: create an object of type int having value 1 and bind the name a to it. The object is an instance of int having value 1, and the name a refers to it. The name a and the object to which it refers are distinct.

Now lets say you do

a += 1

Since ints are immutable, what happens here is as follows:

1. look up the object that a refers to (it is an int with id 0x559239eeb380)
2. look up the value of object 0x559239eeb380 (it is 1)
3. add 1 to that value (1 + 1 = 2)
4. create a new int object with value 2 (it has object id 0x559239eeb3a0)
5. rebind the name a to this new object
6. Now a refers to object 0x559239eeb3a0 and the original object (0x559239eeb380) is no longer refered to by the name a. If there aren't any other names refering to the original object it will be garbage collected later.

Give it a try yourself:

a = 1
print(hex(id(a)))
a += 1
print(hex(id(a)))

Answer 9

Yeah, I missed ++ and -- functionality as well. A few million lines of c code engrained that kind of thinking in my old head, and rather than fight it... Here's a class I cobbled up that implements:

pre- and post-increment, pre- and post-decrement, addition,
subtraction, multiplication, division, results assignable
as integer, printable, settable.

Here 'tis:

class counter(object):
    def __init__(self,v=0):
        self.set(v)

    def preinc(self):
        self.v += 1
        return self.v
    def predec(self):
        self.v -= 1
        return self.v

    def postinc(self):
        self.v += 1
        return self.v - 1
    def postdec(self):
        self.v -= 1
        return self.v + 1

    def __add__(self,addend):
        return self.v + addend
    def __sub__(self,subtrahend):
        return self.v - subtrahend
    def __mul__(self,multiplier):
        return self.v * multiplier
    def __div__(self,divisor):
        return self.v / divisor

    def __getitem__(self):
        return self.v

    def __str__(self):
        return str(self.v)

    def set(self,v):
        if type(v) != int:
            v = 0
        self.v = v

You might use it like this:

c = counter()                          # defaults to zero
for listItem in myList:                # imaginary task
     doSomething(c.postinc(),listItem) # passes c, but becomes c+1

...already having c, you could do this...

c.set(11)
while c.predec() > 0:
    print c

....or just...

d = counter(11)
while d.predec() > 0:
    print d

...and for (re-)assignment into integer...

c = counter(100)
d = c + 223 # assignment as integer
c = c + 223 # re-assignment as integer
print type(c),c # <type 'int'> 323

...while this will maintain c as type counter:

c = counter(100)
c.set(c + 223)
print type(c),c # <class '__main__.counter'> 323

EDIT:

And then there's this bit of unexpected (and thoroughly unwanted) behavior,

c = counter(42)
s = '%s: %d' % ('Expecting 42',c) # but getting non-numeric exception
print s

...because inside that tuple, getitem() isn't what used, instead a reference to the object is passed to the formatting function. Sigh. So:

c = counter(42)
s = '%s: %d' % ('Expecting 42',c.v) # and getting 42.
print s

...or, more verbosely, and explicitly what we actually wanted to happen, although counter-indicated in actual form by the verbosity (use c.v instead)...

c = counter(42)
s = '%s: %d' % ('Expecting 42',c.__getitem__()) # and getting 42.
print s

Answer 10

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.

Answer 11

In Python, a distinction between expressions and statements is rigidly enforced, in contrast to languages such as Common Lisp, Scheme, or Ruby.

Wikipedia

So by introducing such operators, you would break the expression/statement split.

For the same reason you can't write

if x = 0:
  y = 1

as you can in some other languages where such distinction is not preserved.