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How does Python know the values already stored in its memory?
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I want to know how Python knows (if it knows) if a value-type object is already stored in its memory (and also knows where it is).
For this code, when assigning the value 1
for b
, how does it know that the value 1
is already in its memory and stores its reference in b
?
>>> a = 1
>>> b = 1
>>> a is b
True
python python-3.x memory
New contributor
|
show 1 more comment
I want to know how Python knows (if it knows) if a value-type object is already stored in its memory (and also knows where it is).
For this code, when assigning the value 1
for b
, how does it know that the value 1
is already in its memory and stores its reference in b
?
>>> a = 1
>>> b = 1
>>> a is b
True
python python-3.x memory
New contributor
Useprint(hex(id(b)))
to check memory address forb
– Yusufsn
45 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
the values are the same
– Just A Lone
42 mins ago
1
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
1
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago
|
show 1 more comment
I want to know how Python knows (if it knows) if a value-type object is already stored in its memory (and also knows where it is).
For this code, when assigning the value 1
for b
, how does it know that the value 1
is already in its memory and stores its reference in b
?
>>> a = 1
>>> b = 1
>>> a is b
True
python python-3.x memory
New contributor
I want to know how Python knows (if it knows) if a value-type object is already stored in its memory (and also knows where it is).
For this code, when assigning the value 1
for b
, how does it know that the value 1
is already in its memory and stores its reference in b
?
>>> a = 1
>>> b = 1
>>> a is b
True
python python-3.x memory
python python-3.x memory
New contributor
New contributor
edited 32 mins ago
user11206537
151115
151115
New contributor
asked 49 mins ago
Just A LoneJust A Lone
434
434
New contributor
New contributor
Useprint(hex(id(b)))
to check memory address forb
– Yusufsn
45 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
the values are the same
– Just A Lone
42 mins ago
1
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
1
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago
|
show 1 more comment
Useprint(hex(id(b)))
to check memory address forb
– Yusufsn
45 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
the values are the same
– Just A Lone
42 mins ago
1
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
1
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago
Use
print(hex(id(b)))
to check memory address for b
– Yusufsn
45 mins ago
Use
print(hex(id(b)))
to check memory address for b
– Yusufsn
45 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
the values are the same
– Just A Lone
42 mins ago
the values are the same
– Just A Lone
42 mins ago
1
1
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
1
1
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago
|
show 1 more comment
4 Answers
4
active
oldest
votes
Python uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.
a = 10e5
b = 10e5
a is b # False
Wait, what? If you check the address of the numbers, you'll find something interesting:
a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512
a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712
It's called integer cache. You can read more about the integer cache here: https://wsvincent.com/python-wat-integer-cache/
“The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object.”
Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.
New contributor
add a comment |
If you take a look at Objects/longobject.c
, which implements the int
type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS
) and 256 (NSMALLPOSINTS - 1
) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.
add a comment |
Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.
New contributor
I think you're missing the point of the question.a == b
is obviously true. OP is asking whya is b
is true.
– Mad Physicist
43 mins ago
add a comment |
Why?
is
is kinda confusing:
@ajnLJA-0184 is correct, but additionally, strings work well too, but when you do some operations to it, but still the same strings, nope:
>>> a = 'python'
>>> b = 'python'
>>> a is b
True
>>> a = 'pytho' + 'n'
>>> b = 'pythonn'[:-1]
>>> a is b
False
>>>
But strange enough, it is different for integers:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> a = 3*8
>>> b = 4*6
>>> a is b
True
>>>
But as the link he gave, there you can see that:
>>> a = 257
>>> b = 257
>>> a is b
False
>>>
How to know if it is gonna be True
or not?
Well, here is when id
comes in handy:
Here you go, just type in id
and two parens and say your varaible...:
>>> a = 1
>>> b = 1
>>> id(a)
1935522256
>>> id(b)
1935522256
>>> a = 257 # or -6
>>> b = 257 # or -6
>>> id(a)
935705330960
>>> id(b)
935705331216
>>>
1
Whyd is a
equals False in the first example? That's new.
– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like'python'
, they do an operation to get'python'
, that's why.
– U9-Forward
24 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Python uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.
a = 10e5
b = 10e5
a is b # False
Wait, what? If you check the address of the numbers, you'll find something interesting:
a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512
a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712
It's called integer cache. You can read more about the integer cache here: https://wsvincent.com/python-wat-integer-cache/
“The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object.”
Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.
New contributor
add a comment |
Python uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.
a = 10e5
b = 10e5
a is b # False
Wait, what? If you check the address of the numbers, you'll find something interesting:
a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512
a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712
It's called integer cache. You can read more about the integer cache here: https://wsvincent.com/python-wat-integer-cache/
“The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object.”
Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.
New contributor
add a comment |
Python uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.
a = 10e5
b = 10e5
a is b # False
Wait, what? If you check the address of the numbers, you'll find something interesting:
a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512
a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712
It's called integer cache. You can read more about the integer cache here: https://wsvincent.com/python-wat-integer-cache/
“The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object.”
Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.
New contributor
Python uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.
a = 10e5
b = 10e5
a is b # False
Wait, what? If you check the address of the numbers, you'll find something interesting:
a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512
a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712
It's called integer cache. You can read more about the integer cache here: https://wsvincent.com/python-wat-integer-cache/
“The current implementation keeps an array of integer objects for all
integers between -5 and 256, when you create an int in that range you
actually just get back a reference to the existing object.”
Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.
New contributor
edited 38 mins ago
New contributor
answered 44 mins ago
ajnLJA-0184ajnLJA-0184
1984
1984
New contributor
New contributor
add a comment |
add a comment |
If you take a look at Objects/longobject.c
, which implements the int
type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS
) and 256 (NSMALLPOSINTS - 1
) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.
add a comment |
If you take a look at Objects/longobject.c
, which implements the int
type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS
) and 256 (NSMALLPOSINTS - 1
) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.
add a comment |
If you take a look at Objects/longobject.c
, which implements the int
type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS
) and 256 (NSMALLPOSINTS - 1
) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.
If you take a look at Objects/longobject.c
, which implements the int
type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS
) and 256 (NSMALLPOSINTS - 1
) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.
answered 38 mins ago
Mad PhysicistMad Physicist
39k1682113
39k1682113
add a comment |
add a comment |
Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.
New contributor
I think you're missing the point of the question.a == b
is obviously true. OP is asking whya is b
is true.
– Mad Physicist
43 mins ago
add a comment |
Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.
New contributor
I think you're missing the point of the question.a == b
is obviously true. OP is asking whya is b
is true.
– Mad Physicist
43 mins ago
add a comment |
Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.
New contributor
Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.
New contributor
New contributor
answered 44 mins ago
Monster AR44Monster AR44
111
111
New contributor
New contributor
I think you're missing the point of the question.a == b
is obviously true. OP is asking whya is b
is true.
– Mad Physicist
43 mins ago
add a comment |
I think you're missing the point of the question.a == b
is obviously true. OP is asking whya is b
is true.
– Mad Physicist
43 mins ago
I think you're missing the point of the question.
a == b
is obviously true. OP is asking why a is b
is true.– Mad Physicist
43 mins ago
I think you're missing the point of the question.
a == b
is obviously true. OP is asking why a is b
is true.– Mad Physicist
43 mins ago
add a comment |
Why?
is
is kinda confusing:
@ajnLJA-0184 is correct, but additionally, strings work well too, but when you do some operations to it, but still the same strings, nope:
>>> a = 'python'
>>> b = 'python'
>>> a is b
True
>>> a = 'pytho' + 'n'
>>> b = 'pythonn'[:-1]
>>> a is b
False
>>>
But strange enough, it is different for integers:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> a = 3*8
>>> b = 4*6
>>> a is b
True
>>>
But as the link he gave, there you can see that:
>>> a = 257
>>> b = 257
>>> a is b
False
>>>
How to know if it is gonna be True
or not?
Well, here is when id
comes in handy:
Here you go, just type in id
and two parens and say your varaible...:
>>> a = 1
>>> b = 1
>>> id(a)
1935522256
>>> id(b)
1935522256
>>> a = 257 # or -6
>>> b = 257 # or -6
>>> id(a)
935705330960
>>> id(b)
935705331216
>>>
1
Whyd is a
equals False in the first example? That's new.
– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like'python'
, they do an operation to get'python'
, that's why.
– U9-Forward
24 mins ago
add a comment |
Why?
is
is kinda confusing:
@ajnLJA-0184 is correct, but additionally, strings work well too, but when you do some operations to it, but still the same strings, nope:
>>> a = 'python'
>>> b = 'python'
>>> a is b
True
>>> a = 'pytho' + 'n'
>>> b = 'pythonn'[:-1]
>>> a is b
False
>>>
But strange enough, it is different for integers:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> a = 3*8
>>> b = 4*6
>>> a is b
True
>>>
But as the link he gave, there you can see that:
>>> a = 257
>>> b = 257
>>> a is b
False
>>>
How to know if it is gonna be True
or not?
Well, here is when id
comes in handy:
Here you go, just type in id
and two parens and say your varaible...:
>>> a = 1
>>> b = 1
>>> id(a)
1935522256
>>> id(b)
1935522256
>>> a = 257 # or -6
>>> b = 257 # or -6
>>> id(a)
935705330960
>>> id(b)
935705331216
>>>
1
Whyd is a
equals False in the first example? That's new.
– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like'python'
, they do an operation to get'python'
, that's why.
– U9-Forward
24 mins ago
add a comment |
Why?
is
is kinda confusing:
@ajnLJA-0184 is correct, but additionally, strings work well too, but when you do some operations to it, but still the same strings, nope:
>>> a = 'python'
>>> b = 'python'
>>> a is b
True
>>> a = 'pytho' + 'n'
>>> b = 'pythonn'[:-1]
>>> a is b
False
>>>
But strange enough, it is different for integers:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> a = 3*8
>>> b = 4*6
>>> a is b
True
>>>
But as the link he gave, there you can see that:
>>> a = 257
>>> b = 257
>>> a is b
False
>>>
How to know if it is gonna be True
or not?
Well, here is when id
comes in handy:
Here you go, just type in id
and two parens and say your varaible...:
>>> a = 1
>>> b = 1
>>> id(a)
1935522256
>>> id(b)
1935522256
>>> a = 257 # or -6
>>> b = 257 # or -6
>>> id(a)
935705330960
>>> id(b)
935705331216
>>>
Why?
is
is kinda confusing:
@ajnLJA-0184 is correct, but additionally, strings work well too, but when you do some operations to it, but still the same strings, nope:
>>> a = 'python'
>>> b = 'python'
>>> a is b
True
>>> a = 'pytho' + 'n'
>>> b = 'pythonn'[:-1]
>>> a is b
False
>>>
But strange enough, it is different for integers:
>>> a = 1
>>> b = 1
>>> a is b
True
>>> a = 3*8
>>> b = 4*6
>>> a is b
True
>>>
But as the link he gave, there you can see that:
>>> a = 257
>>> b = 257
>>> a is b
False
>>>
How to know if it is gonna be True
or not?
Well, here is when id
comes in handy:
Here you go, just type in id
and two parens and say your varaible...:
>>> a = 1
>>> b = 1
>>> id(a)
1935522256
>>> id(b)
1935522256
>>> a = 257 # or -6
>>> b = 257 # or -6
>>> id(a)
935705330960
>>> id(b)
935705331216
>>>
edited 24 mins ago
answered 34 mins ago
U9-ForwardU9-Forward
18.5k51744
18.5k51744
1
Whyd is a
equals False in the first example? That's new.
– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like'python'
, they do an operation to get'python'
, that's why.
– U9-Forward
24 mins ago
add a comment |
1
Whyd is a
equals False in the first example? That's new.
– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like'python'
, they do an operation to get'python'
, that's why.
– U9-Forward
24 mins ago
1
1
Why
d is a
equals False in the first example? That's new.– ajnLJA-0184
26 mins ago
Why
d is a
equals False in the first example? That's new.– ajnLJA-0184
26 mins ago
@ajnLJA-0 Because the strings aren't direct like
'python'
, they do an operation to get 'python'
, that's why.– U9-Forward
24 mins ago
@ajnLJA-0 Because the strings aren't direct like
'python'
, they do an operation to get 'python'
, that's why.– U9-Forward
24 mins ago
add a comment |
Just A Lone is a new contributor. Be nice, and check out our Code of Conduct.
Just A Lone is a new contributor. Be nice, and check out our Code of Conduct.
Just A Lone is a new contributor. Be nice, and check out our Code of Conduct.
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Use
print(hex(id(b)))
to check memory address forb
– Yusufsn
45 mins ago
>>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350'
– Just A Lone
43 mins ago
the values are the same
– Just A Lone
42 mins ago
1
If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object.
– Yusufsn
42 mins ago
1
@Yusufsn No. For bigger integers (>256) it's not true.
– ajnLJA-0184
37 mins ago