Why does calling Python's 'magic method' not do type conversion like it would for the corresponding operator?
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
asked 4 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
asked 4 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
asked 4 hours ago
dopplerdoppler
1316
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
python casting
asked 4 hours ago
dopplerdoppler
1316
asked 4 hours ago
dopplerdoppler
1316
asked 4 hours ago
dopplerdoppler
1316
asked 4 hours ago
dopplerdoppler
1316
asked 4 hours ago
dopplerdoppler
1316
1316
add a comment |
add a comment |
1 Answer
1
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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.
answered 4 hours ago
user2357112user2357112
155k12163258
5
It's worth noting that theNotImplementedresult that is returned by the calls in the question are the signal to try the reverse method.
– Blckknght
4 hours ago
Thanks! I was aware it would try__rsub__but didn't know it would reverse the argument order.
– doppler
3 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
3 hours ago
@user2357112 soself.__rsub__(other)really just callsother.__sub__(self), if that makes any sense?
– doppler
3 hours ago
2
@doppler: No.self.__rsub__(other)is called forother - selfifothercan't handle it. Callingother.__sub__(self)would be pointless. We already knowothercan't handle it.
– user2357112
3 hours ago
|
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1 Answer
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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.
answered 4 hours ago
user2357112user2357112
155k12163258
5
It's worth noting that theNotImplementedresult that is returned by the calls in the question are the signal to try the reverse method.
– Blckknght
4 hours ago
Thanks! I was aware it would try__rsub__but didn't know it would reverse the argument order.
– doppler
3 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
3 hours ago
@user2357112 soself.__rsub__(other)really just callsother.__sub__(self), if that makes any sense?
– doppler
3 hours ago
2
@doppler: No.self.__rsub__(other)is called forother - selfifothercan't handle it. Callingother.__sub__(self)would be pointless. We already knowothercan't handle it.
– user2357112
3 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.
answered 4 hours ago
user2357112user2357112
155k12163258
5
It's worth noting that theNotImplementedresult that is returned by the calls in the question are the signal to try the reverse method.
– Blckknght
4 hours ago
Thanks! I was aware it would try__rsub__but didn't know it would reverse the argument order.
– doppler
3 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
3 hours ago
@user2357112 soself.__rsub__(other)really just callsother.__sub__(self), if that makes any sense?
– doppler
3 hours ago
2
@doppler: No.self.__rsub__(other)is called forother - selfifothercan't handle it. Callingother.__sub__(self)would be pointless. We already knowothercan't handle it.
– user2357112
3 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.
answered 4 hours ago
user2357112user2357112
155k12163258
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.
answered 4 hours ago
user2357112user2357112
155k12163258
edited 4 hours ago
answered 4 hours ago
user2357112user2357112
155k12163258
answered 4 hours ago
user2357112user2357112
155k12163258
answered 4 hours ago
user2357112user2357112
155k12163258
155k12163258
5
It's worth noting that theNotImplementedresult that is returned by the calls in the question are the signal to try the reverse method.
– Blckknght
4 hours ago
Thanks! I was aware it would try__rsub__but didn't know it would reverse the argument order.
– doppler
3 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
3 hours ago
@user2357112 soself.__rsub__(other)really just callsother.__sub__(self), if that makes any sense?
– doppler
3 hours ago
2
@doppler: No.self.__rsub__(other)is called forother - selfifothercan't handle it. Callingother.__sub__(self)would be pointless. We already knowothercan't handle it.
– user2357112
3 hours ago
|
show 1 more comment
5
It's worth noting that theNotImplementedresult that is returned by the calls in the question are the signal to try the reverse method.
– Blckknght
4 hours ago
Thanks! I was aware it would try__rsub__but didn't know it would reverse the argument order.
– doppler
3 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
3 hours ago
@user2357112 soself.__rsub__(other)really just callsother.__sub__(self), if that makes any sense?
– doppler
3 hours ago
2
@doppler: No.self.__rsub__(other)is called forother - selfifothercan't handle it. Callingother.__sub__(self)would be pointless. We already knowothercan't handle it.
– user2357112
3 hours ago
5
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
4 hours ago
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
4 hours ago
Thanks! I was aware it would try
__rsub__ but didn't know it would reverse the argument order.– doppler
3 hours ago
Thanks! I was aware it would try
__rsub__ but didn't know it would reverse the argument order.– doppler
3 hours ago
1
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
3 hours ago
@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
3 hours ago
@user2357112 so
self.__rsub__(other) really just calls other.__sub__(self), if that makes any sense?– doppler
3 hours ago
@user2357112 so
self.__rsub__(other) really just calls other.__sub__(self), if that makes any sense?– doppler
3 hours ago
2
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
3 hours ago
@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
3 hours ago
|
show 1 more comment
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