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Why does calling Python's 'magic method' not do type conversion like it would for the corresponding operator?

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8

When I subtract a float from an integer (e.g. 1-2.0), Python does implicit type conversion (I think). But when I call what I thought was the same operation using the magic method __sub__, it suddenly does not anymore.

What am I missing here? When I overload operators for my own classes, is there a way around this other than explicitly casting input to whatever type I need?

a=1

a.__sub__(2.)

# returns NotImplemented

a.__rsub__(2.)

# returns NotImplemented

# yet, of course:

a-2.

# returns -1.0

python casting

share|improve this question

asked 3 hours ago

dopplerdoppler

1316

add a comment | 

8

When I subtract a float from an integer (e.g. 1-2.0), Python does implicit type conversion (I think). But when I call what I thought was the same operation using the magic method __sub__, it suddenly does not anymore.

What am I missing here? When I overload operators for my own classes, is there a way around this other than explicitly casting input to whatever type I need?

a=1

a.__sub__(2.)

# returns NotImplemented

a.__rsub__(2.)

# returns NotImplemented

# yet, of course:

a-2.

# returns -1.0

python casting

share|improve this question

asked 3 hours ago

dopplerdoppler

1316

add a comment | 

When I subtract a float from an integer (e.g. 1-2.0), Python does implicit type conversion (I think). But when I call what I thought was the same operation using the magic method __sub__, it suddenly does not anymore.

What am I missing here? When I overload operators for my own classes, is there a way around this other than explicitly casting input to whatever type I need?

a=1

a.__sub__(2.)

# returns NotImplemented

a.__rsub__(2.)

# returns NotImplemented

# yet, of course:

a-2.

# returns -1.0

python casting

share|improve this question

asked 3 hours ago

dopplerdoppler

1316

a=1

a.__sub__(2.)

# returns NotImplemented

a.__rsub__(2.)

# returns NotImplemented

# yet, of course:

a-2.

# returns -1.0

share|improve this question

asked 3 hours ago

dopplerdoppler

1316

share|improve this question

asked 3 hours ago

dopplerdoppler

1316

asked 3 hours ago

dopplerdoppler

1316

asked 3 hours ago

dopplerdoppler

1316

1 Answer
1

active

oldest

votes

14

a - b isn't just a.__sub__(b). It also tries b.__rsub__(a) if a can't handle the operation, and in the 1 - 2. case, it's the float's __rsub__ that handles the operation.

>>> (2.).__rsub__(1)

-1.0

You ran a.__rsub__(2.), but that's the wrong __rsub__. You need the right-side operand's __rsub__, not the left-side operand.

---

There is no implicit type conversion built into the subtraction operator. float.__rsub__ has to handle ints manually. If you want type conversion in your own operator implementations, you'll have to handle that manually too.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

user2357112user2357112

155k12163258

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  It's worth noting that the NotImplemented result that is returned by the calls in the question are the signal to try the reverse method.
  
  – Blckknght
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I was aware it would try __rsub__ but didn't know it would reverse the argument order.
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @doppler: It'd be pretty pointless to have the left operand handle both __sub__ and __rsub__. That'd just be two methods with the exact same job, and the right operand would have no opportunity to supply an implementation.
  
  – user2357112
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @user2357112 so self.__rsub__(other) really just calls other.__sub__(self), if that makes any sense?
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @doppler: No. self.__rsub__(other) is called for other - self if other can't handle it. Calling other.__sub__(self) would be pointless. We already know other can't handle it.
  
  – user2357112
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

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14

a - b isn't just a.__sub__(b). It also tries b.__rsub__(a) if a can't handle the operation, and in the 1 - 2. case, it's the float's __rsub__ that handles the operation.

>>> (2.).__rsub__(1)

-1.0

You ran a.__rsub__(2.), but that's the wrong __rsub__. You need the right-side operand's __rsub__, not the left-side operand.

---

There is no implicit type conversion built into the subtraction operator. float.__rsub__ has to handle ints manually. If you want type conversion in your own operator implementations, you'll have to handle that manually too.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

user2357112user2357112

155k12163258

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  It's worth noting that the NotImplemented result that is returned by the calls in the question are the signal to try the reverse method.
  
  – Blckknght
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I was aware it would try __rsub__ but didn't know it would reverse the argument order.
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @doppler: It'd be pretty pointless to have the left operand handle both __sub__ and __rsub__. That'd just be two methods with the exact same job, and the right operand would have no opportunity to supply an implementation.
  
  – user2357112
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @user2357112 so self.__rsub__(other) really just calls other.__sub__(self), if that makes any sense?
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @doppler: No. self.__rsub__(other) is called for other - self if other can't handle it. Calling other.__sub__(self) would be pointless. We already know other can't handle it.
  
  – user2357112
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

14

a - b isn't just a.__sub__(b). It also tries b.__rsub__(a) if a can't handle the operation, and in the 1 - 2. case, it's the float's __rsub__ that handles the operation.

>>> (2.).__rsub__(1)

-1.0

You ran a.__rsub__(2.), but that's the wrong __rsub__. You need the right-side operand's __rsub__, not the left-side operand.

---

There is no implicit type conversion built into the subtraction operator. float.__rsub__ has to handle ints manually. If you want type conversion in your own operator implementations, you'll have to handle that manually too.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

user2357112user2357112

155k12163258

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  It's worth noting that the NotImplemented result that is returned by the calls in the question are the signal to try the reverse method.
  
  – Blckknght
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks! I was aware it would try __rsub__ but didn't know it would reverse the argument order.
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @doppler: It'd be pretty pointless to have the left operand handle both __sub__ and __rsub__. That'd just be two methods with the exact same job, and the right operand would have no opportunity to supply an implementation.
  
  – user2357112
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @user2357112 so self.__rsub__(other) really just calls other.__sub__(self), if that makes any sense?
  
  – doppler
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @doppler: No. self.__rsub__(other) is called for other - self if other can't handle it. Calling other.__sub__(self) would be pointless. We already know other can't handle it.
  
  – user2357112
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

a - b isn't just a.__sub__(b). It also tries b.__rsub__(a) if a can't handle the operation, and in the 1 - 2. case, it's the float's __rsub__ that handles the operation.

>>> (2.).__rsub__(1)

-1.0

You ran a.__rsub__(2.), but that's the wrong __rsub__. You need the right-side operand's __rsub__, not the left-side operand.

---

There is no implicit type conversion built into the subtraction operator. float.__rsub__ has to handle ints manually. If you want type conversion in your own operator implementations, you'll have to handle that manually too.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

user2357112user2357112

155k12163258

>>> (2.).__rsub__(1)

-1.0

share|improve this answer

edited 2 hours ago

answered 2 hours ago

user2357112user2357112

155k12163258

answered 2 hours ago

user2357112user2357112

155k12163258

answered 2 hours ago

user2357112user2357112

155k12163258