Hi there,
I'm not sure, if you would still be interested? :)
- aatree.cpp, lines 171+
- In line 171 you store
t->right in variable m
- In line 172
Skew(m) might return a reference that may be different from m
- So line 172 may change
t->right, but m remains as is
- In line 173 you try to use
m to test if t->right is NULL or not, even though m might no longer be the same as the current t->right
Are you sure that isn't a bug?
Hi there,
I'm not sure, if you would still be interested? :)
t->rightin variablemSkew(m)might return a reference that may be different frommt->right, butmremains as ismto test ift->rightisNULLor not, even thoughmmight no longer be the same as the currentt->rightAre you sure that isn't a bug?