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How to compare two methods? #863
Description
Hi.
I want to check whether a given callback is a specific method in a class. I found that comparing two methods failed. Following is the test code:
class ClassA:
def proc(self, callback=None):
if callback==self.method1:
print('callback==self.method1, callback=', callback)
else:
print('callback!=self.method1')
print('callback.name=', callback.name)
print('callback.__name__=', callback.__name__)
print('callback.__qualname__=', callback.__qualname__)
print('callback=', callback)
print('self.method1=', self.method1)
if self.method1==self.method1:
print('self.method1==self.method1')
else: print('self.method1!=self.method1')
def method1(self):
pass
def main():
a1= ClassA()
a1.proc(a1.method1)
if __name__ == '__main__':
main()
output:
callback!=self.method1
callback.name= None
callback.__name__= None
callback.__qualname__= None
callback= function () {
var args = [] .slice.apply (arguments);
return func.apply (null, [self] .concat (args));
}
self.method1= function () {
var args = [] .slice.apply (arguments);
return func.apply (null, [self] .concat (args));
}
self.method1!=self.method1
What I was surprised was the last line in the output, self.method1!=self.method1, which was a self comparison, but failed. (Is it a bug?)
How can I know a given callback is equal to a specific method? I tried to find a workaround. But as the output above, information about the method (name, _ _ name _ _, and _ _ qualname _ _) is not available. Any idea or workaround is welcome if you have any.
Thanks.
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