Why does str(A())
seemingly call A.__repr__()
and not dict.__str__()
in the example below?
class A(dict):
def __repr__(self):
return 'repr(A)'
def __str__(self):
return dict.__str__(self)
class B(dict):
def __str__(self):
return dict.__str__(self)
print 'call: repr(A) expect: repr(A) get:', repr(A()) # works
print 'call: str(A) expect: {} get:', str(A()) # does not work
print 'call: str(B) expect: {} get:', str(B()) # works
Output:
call: repr(A) expect: repr(A) get: repr(A)
call: str(A) expect: {} get: repr(A)
call: str(B) expect: {} get: {}
str(A())
does call __str__
, in turn calling dict.__str__()
.
It is dict.__str__()
that returns the value repr(A).
I have modified the code to clear things out:
class A(dict):
def __repr__(self):
print "repr of A called",
return 'repr(A)'
def __str__(self):
print "str of A called",
return dict.__str__(self)
class B(dict):
def __str__(self):
print "str of B called",
return dict.__str__(self)
And the output is:
>>> print 'call: repr(A) expect: repr(A) get:', repr(A())
call: repr(A) expect: repr(A) get: repr of A called repr(A)
>>> print 'call: str(A) expect: {} get:', str(A())
call: str(A) expect: {} get: str of A called repr of A called repr(A)
>>> print 'call: str(B) expect: {} get:', str(B())
call: str(B) expect: {} get: str of B called {}
Meaning that str function calls the repr function automatically. And since it was redefined with class A, it returns the 'unexpected' value.
I have posted a workaround solution to it. Check it out ... you might find it useful: http://blog.teltub.com/2009/10/workaround-solution-to-python-strrepr.html
PS Read the original post where I introduced the issue as well ... There problem is the unexpected behavior that catches you by surprise ...
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