When filling in a form, the user needs to specify an amount. This amount is then checked against approximately 4 to 6 ranges. The selected range is then saved in the database. The original amount will not be stored (for non-technical reasons). There will be no overlay between the ranges, e.g.:
0-999
1000-1999
2000-4999
5000-9999
10000-higher
The tricky part is that these ranges are not fixed in stone. There can be alterations and additional ranges can be added to further specify the '10000 and higher' range. These changes will occur a couple of times and can't be prevented. The old ranges will need to be stored since the specific amount can not be saved to the database.
What would be the most efficient C# data structure for checking against a changing set of ranges?
For my research I included:
One of the answers here suggest that a fixed set of integer ranges in a switch statement is possible with C#7. However, it is not possible to dynamically add cases to and/or remove cases from a switch statement.
This question suggests that using Enumerable.Range is not the most efficient way.
A simple approach here is to store the lower band values in an array, and pass it to a FindBand() method which returns an integer representing the index of the band containing the value.
For example:
public static int FindBand(double value, double[] bandLowerValues)
{
for (int i = 0; i < bandLowerValues.Length; ++i)
if (value < bandLowerValues[i])
return Math.Max(0, i-1);
return bandLowerValues.Length;
}
Test code:
double[] bandLowerValues = {0, 1, 2, 5, 10};
Console.WriteLine(FindBand(-1, bandLowerValues));
Console.WriteLine(FindBand(0, bandLowerValues));
Console.WriteLine(FindBand(0.5, bandLowerValues));
Console.WriteLine(FindBand(1, bandLowerValues));
Console.WriteLine(FindBand(1.5, bandLowerValues));
Console.WriteLine(FindBand(2.5, bandLowerValues));
Console.WriteLine(FindBand(5, bandLowerValues));
Console.WriteLine(FindBand(8, bandLowerValues));
Console.WriteLine(FindBand(9.9, bandLowerValues));
Console.WriteLine(FindBand(10, bandLowerValues));
Console.WriteLine(FindBand(11, bandLowerValues));
This isn't the fastest approach if there are a LOT of bands, but if there are just a few bands this is likely to be sufficiently fast.
(If there were a lot of bands, you could use a binary search to find the appropriate band, but that would be overkill for this in my opinion.)
You can sort low bounds, e.g.
// or decimal instead of double if values are money
double[] lowBounds = new double[] {
0, // 0th group: (-Inf .. 0)
1000, // 1st group: [0 .. 1000)
2000, // 2nd group: [1000 .. 2000)
5000, // 3d group: [2000 .. 5000)
10000, // 4th group: [5000 .. 10000)
// 5th group: [10000 .. +Inf)
};
and then find the correct group (0-based)
int index = Array.BinarySearch(lowBounds, value);
index = index < 0 ? index = -index - 1 : index + 1;
Demo:
double[] tests = new double[] {
-10,
0,
45,
999,
1000,
1997,
5123,
10000,
20000,
};
var result = tests
.Select(value => {
int index = Array.BinarySearch(lowBounds, value);
index = index < 0 ? index = -index - 1 : index + 1;
return $"{value,6} : {index}";
});
Console.Write(string.Join(Environment.NewLine, result));
Outcome:
-10 : 0
0 : 1
45 : 1
999 : 1
1000 : 2
1997 : 2
5123 : 4
10000 : 5
20000 : 5
Since there are already great answers regarding how to find the correct range, I'd like to address the persistence issue.
What do we have here?
You cannot persist the exact value. ( Not allowed )
Values will be "blurred" by fitting them into a range.
Those ranges can (and will) change over time in bounds and number.
So, what I would probably do would be to persist lower and upper bound explicitly in the db.
That way, if ranges change, old data is still correct. You cannot "transform" to the new ranges, because you cannot know if it would be correct. So you need to keep the old values. Any new entries after the change will reflect the new ranges.
One could think of normalization, but honestly, I think that would be overcomplicating the problem. I'd only consider that if the benefit (less storage space) would greatly outweigh the complexity issues.
I would like to be able to get the name of a variable as a string but I don't know if Python has that much introspection capabilities. Something like:
>>> print(my_var.__name__)
'my_var'
I want to do that because I have a bunch of variables I'd like to turn into a dictionary like :
bar = True
foo = False
>>> my_dict = dict(bar=bar, foo=foo)
>>> print my_dict
{'foo': False, 'bar': True}
But I'd like something more automatic than that.
Python have locals() and vars(), so I guess there is a way.
As unwind said, this isn't really something you do in Python - variables are actually name mappings to objects.
However, here's one way to try and do it:
>>> a = 1
>>> for k, v in list(locals().iteritems()):
if v is a:
a_as_str = k
>>> a_as_str
a
>>> type(a_as_str)
'str'
I've wanted to do this quite a lot. This hack is very similar to rlotun's suggestion, but it's a one-liner, which is important to me:
blah = 1
blah_name = [ k for k,v in locals().iteritems() if v is blah][0]
Python 3+
blah = 1
blah_name = [ k for k,v in locals().items() if v is blah][0]
Are you trying to do this?
dict( (name,eval(name)) for name in ['some','list','of','vars'] )
Example
>>> some= 1
>>> list= 2
>>> of= 3
>>> vars= 4
>>> dict( (name,eval(name)) for name in ['some','list','of','vars'] )
{'list': 2, 'some': 1, 'vars': 4, 'of': 3}
This is a hack. It will not work on all Python implementations distributions (in particular, those that do not have traceback.extract_stack.)
import traceback
def make_dict(*expr):
(filename,line_number,function_name,text)=traceback.extract_stack()[-2]
begin=text.find('make_dict(')+len('make_dict(')
end=text.find(')',begin)
text=[name.strip() for name in text[begin:end].split(',')]
return dict(zip(text,expr))
bar=True
foo=False
print(make_dict(bar,foo))
# {'foo': False, 'bar': True}
Note that this hack is fragile:
make_dict(bar,
foo)
(calling make_dict on 2 lines) will not work.
Instead of trying to generate the dict out of the values foo and bar,
it would be much more Pythonic to generate the dict out of the string variable names 'foo' and 'bar':
dict([(name,locals()[name]) for name in ('foo','bar')])
This is not possible in Python, which really doesn't have "variables". Python has names, and there can be more than one name for the same object.
I think my problem will help illustrate why this question is useful, and it may give a bit more insight into how to answer it. I wrote a small function to do a quick inline head check on various variables in my code. Basically, it lists the variable name, data type, size, and other attributes, so I can quickly catch any mistakes I've made. The code is simple:
def details(val):
vn = val.__name__ # If such a thing existed
vs = str(val)
print("The Value of "+ str(vn) + " is " + vs)
print("The data type of " + vn + " is " + str(type(val)))
So if you have some complicated dictionary / list / tuple situation, it would be quite helpful to have the interpreter return the variable name you assigned. For instance, here is a weird dictionary:
m = 'abracadabra'
mm=[]
for n in m:
mm.append(n)
mydic = {'first':(0,1,2,3,4,5,6),'second':mm,'third':np.arange(0.,10)}
details(mydic)
The Value of mydic is {'second': ['a', 'b', 'r', 'a', 'c', 'a', 'd', 'a', 'b', 'r', 'a'], 'third': array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.]), 'first': [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]}
The data type of mydic is <type 'dict'>
details(mydic['first'])
The Value of mydic['first'] is (0, 1, 2, 3, 4, 5, 6)]
The data type of mydic['first'] is <type 'list'>
details(mydic.keys())
The Value of mydic.keys() is ['second', 'third', 'first']
The data type of mydic.keys() is <type 'tuple'>
details(mydic['second'][0])
The Value of mydic['second'][0] is a
The data type of mydic['second'][0] is <type 'str'>
I'm not sure if I put this in the right place, but I thought it might help. I hope it does.
I wrote a neat little useful function based on the answer to this question. I'm putting it here in case it's useful.
def what(obj, callingLocals=locals()):
"""
quick function to print name of input and value.
If not for the default-Valued callingLocals, the function would always
get the name as "obj", which is not what I want.
"""
for k, v in list(callingLocals.items()):
if v is obj:
name = k
print(name, "=", obj)
usage:
>> a = 4
>> what(a)
a = 4
>>|
I find that if you already have a specific list of values, that the way described by #S. Lotts is the best; however, the way described below works well to get all variables and Classes added throughout the code WITHOUT the need to provide variable name though you can specify them if you want. Code can be extend to exclude Classes.
import types
import math # mainly showing that you could import what you will before d
# Everything after this counts
d = dict(globals())
def kv_test(k,v):
return (k not in d and
k not in ['d','args'] and
type(v) is not types.FunctionType)
def magic_print(*args):
if len(args) == 0:
return {k:v for k,v in globals().iteritems() if kv_test(k,v)}
else:
return {k:v for k,v in magic_print().iteritems() if k in args}
if __name__ == '__main__':
foo = 1
bar = 2
baz = 3
print magic_print()
print magic_print('foo')
print magic_print('foo','bar')
Output:
{'baz': 3, 'foo': 1, 'bar': 2}
{'foo': 1}
{'foo': 1, 'bar': 2}
In python 3 this is easy
myVariable = 5
for v in locals():
if id(v) == id("myVariable"):
print(v, locals()[v])
this will print:
myVariable 5
Python3. Use inspect to capture the calling local namespace then use ideas presented here. Can return more than one answer as has been pointed out.
def varname(var):
import inspect
frame = inspect.currentframe()
var_id = id(var)
for name in frame.f_back.f_locals.keys():
try:
if id(eval(name)) == var_id:
return(name)
except:
pass
Here's the function I created to read the variable names. It's more general and can be used in different applications:
def get_variable_name(*variable):
'''gets string of variable name
inputs
variable (str)
returns
string
'''
if len(variable) != 1:
raise Exception('len of variables inputed must be 1')
try:
return [k for k, v in locals().items() if v is variable[0]][0]
except:
return [k for k, v in globals().items() if v is variable[0]][0]
To use it in the specified question:
>>> foo = False
>>> bar = True
>>> my_dict = {get_variable_name(foo):foo,
get_variable_name(bar):bar}
>>> my_dict
{'bar': True, 'foo': False}
In reading the thread, I saw an awful lot of friction. It's easy enough to give
a bad answer, then let someone give the correct answer. Anyway, here is what I found.
From: [effbot.org] (http://effbot.org/zone/python-objects.htm#names)
The names are a bit different — they’re not really properties of the object, and the object itself doesn't know what it’s called.
An object can have any number of names, or no name at all.
Names live in namespaces (such as a module namespace, an instance namespace, a function’s local namespace).
Note: that it says the object itself doesn’t know what it’s called, so that was the clue. Python objects are not self-referential. Then it says, Names live in namespaces. We have this in TCL/TK. So maybe my answer will help (but it did help me)
jj = 123
print eval("'" + str(id(jj)) + "'")
print dir()
166707048
['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj']
So there is 'jj' at the end of the list.
Rewrite the code as:
jj = 123
print eval("'" + str(id(jj)) + "'")
for x in dir():
print id(eval(x))
161922920
['__builtins__', '__doc__', '__file__', '__name__', '__package__', 'jj']
3077447796
136515736
3077408320
3077656800
136515736
161922920
This nasty bit of code id's the name of variable/object/whatever-you-pedantics-call-it.
So, there it is. The memory address of 'jj' is the same when we look for it directly, as when we do the dictionary look up in global name space. I'm sure you can make a function to do this. Just remember which namespace your variable/object/wypci is in.
QED.
I wrote the package sorcery to do this kind of magic robustly. You can write:
from sorcery import dict_of
my_dict = dict_of(foo, bar)
Maybe I'm overthinking this but..
str_l = next((k for k,v in locals().items() if id(l) == id(v)))
>>> bar = True
>>> foo = False
>>> my_dict=dict(bar=bar, foo=foo)
>>> next((k for k,v in locals().items() if id(bar) == id(v)))
'bar'
>>> next((k for k,v in locals().items() if id(foo) == id(v)))
'foo'
>>> next((k for k,v in locals().items() if id(my_dict) == id(v)))
'my_dict'
import re
import traceback
pattren = re.compile(r'[\W+\w+]*get_variable_name\((\w+)\)')
def get_variable_name(x):
return pattren.match( traceback.extract_stack(limit=2)[0][3]) .group(1)
a = 1
b = a
c = b
print get_variable_name(a)
print get_variable_name(b)
print get_variable_name(c)
I uploaded a solution to pypi. It's a module defining an equivalent of C#'s nameof function.
It iterates through bytecode instructions for the frame its called in, getting the names of variables/attributes passed to it. The names are found in the .argrepr of LOAD instructions following the function's name.
Most objects don't have a __name__ attribute. (Classes, functions, and modules do; any more builtin types that have one?)
What else would you expect for print(my_var.__name__) other than print("my_var")? Can you simply use the string directly?
You could "slice" a dict:
def dict_slice(D, keys, default=None):
return dict((k, D.get(k, default)) for k in keys)
print dict_slice(locals(), ["foo", "bar"])
# or use set literal syntax if you have a recent enough version:
print dict_slice(locals(), {"foo", "bar"})
Alternatively:
throw = object() # sentinel
def dict_slice(D, keys, default=throw):
def get(k):
v = D.get(k, throw)
if v is not throw:
return v
if default is throw:
raise KeyError(k)
return default
return dict((k, get(k)) for k in keys)
Well, I encountered the very same need a few days ago and had to get a variable's name which was pointing to the object itself.
And why was it so necessary?
In short I was building a plug-in for Maya. The core plug-in was built using C++ but the GUI is drawn through Python(as its not processor intensive). Since I, as yet, don't know how to return multiple values from the plug-in except the default MStatus, therefore to update a dictionary in Python I had to pass the the name of the variable, pointing to the object implementing the GUI and which contained the dictionary itself, to the plug-in and then use the MGlobal::executePythonCommand() to update the dictionary from the global scope of Maya.
To do that what I did was something like:
import time
class foo(bar):
def __init__(self):
super(foo, self).__init__()
self.time = time.time() #almost guaranteed to be unique on a single computer
def name(self):
g = globals()
for x in g:
if isinstance(g[x], type(self)):
if g[x].time == self.time:
return x
#or you could:
#return filter(None,[x if g[x].time == self.time else None for x in g if isinstance(g[x], type(self))])
#and return all keys pointing to object itself
I know that it is not the perfect solution in in the globals many keys could be pointing to the same object e.g.:
a = foo()
b = a
b.name()
>>>b
or
>>>a
and that the approach isn't thread-safe. Correct me if I am wrong.
At least this approach solved my problem by getting the name of any variable in the global scope which pointed to the object itself and pass it over to the plug-in, as argument, for it use internally.
I tried this on int (the primitive integer class) but the problem is that these primitive classes don't get bypassed (please correct the technical terminology used if its wrong). You could re-implement int and then do int = foo but a = 3 will never be an object of foo but of the primitive. To overcome that you have to a = foo(3) to get a.name() to work.
With python 2.7 and newer there is also dictionary comprehension which makes it a bit shorter. If possible I would use getattr instead eval (eval is evil) like in the top answer. Self can be any object which has the context your a looking at. It can be an object or locals=locals() etc.
{name: getattr(self, name) for name in ['some', 'vars', 'here]}
I was working on a similar problem. #S.Lott said "If you have the list of variables, what's the point of "discovering" their names?" And my answer is just to see if it could be done and if for some reason you want to sort your variables by type into lists. So anyways, in my research I came came across this thread and my solution is a bit expanded and is based on #rlotun solution. One other thing, #unutbu said, "This idea has merit, but note that if two variable names reference the same value (e.g. True), then an unintended variable name might be returned." In this exercise that was true so I dealt with it by using a list comprehension similar to this for each possibility: isClass = [i for i in isClass if i != 'item']. Without it "item" would show up in each list.
__metaclass__ = type
from types import *
class Class_1: pass
class Class_2: pass
list_1 = [1, 2, 3]
list_2 = ['dog', 'cat', 'bird']
tuple_1 = ('one', 'two', 'three')
tuple_2 = (1000, 2000, 3000)
dict_1 = {'one': 1, 'two': 2, 'three': 3}
dict_2 = {'dog': 'collie', 'cat': 'calico', 'bird': 'robin'}
x = 23
y = 29
pie = 3.14159
eee = 2.71828
house = 'single story'
cabin = 'cozy'
isClass = []; isList = []; isTuple = []; isDict = []; isInt = []; isFloat = []; isString = []; other = []
mixedDataTypes = [Class_1, list_1, tuple_1, dict_1, x, pie, house, Class_2, list_2, tuple_2, dict_2, y, eee, cabin]
print '\nMIXED_DATA_TYPES total count:', len(mixedDataTypes)
for item in mixedDataTypes:
try:
# if isinstance(item, ClassType): # use this for old class types (before 3.0)
if isinstance(item, type):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isClass.append(mapping_as_str)
isClass = [i for i in isClass if i != 'item']
elif isinstance(item, ListType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isList.append(mapping_as_str)
isList = [i for i in isList if i != 'item']
elif isinstance(item, TupleType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isTuple.append(mapping_as_str)
isTuple = [i for i in isTuple if i != 'item']
elif isinstance(item, DictType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isDict.append(mapping_as_str)
isDict = [i for i in isDict if i != 'item']
elif isinstance(item, IntType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isInt.append(mapping_as_str)
isInt = [i for i in isInt if i != 'item']
elif isinstance(item, FloatType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isFloat.append(mapping_as_str)
isFloat = [i for i in isFloat if i != 'item']
elif isinstance(item, StringType):
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
isString.append(mapping_as_str)
isString = [i for i in isString if i != 'item']
else:
for k, v in list(locals().iteritems()):
if v is item:
mapping_as_str = k
other.append(mapping_as_str)
other = [i for i in other if i != 'item']
except (TypeError, AttributeError), e:
print e
print '\n isClass:', len(isClass), isClass
print ' isList:', len(isList), isList
print ' isTuple:', len(isTuple), isTuple
print ' isDict:', len(isDict), isDict
print ' isInt:', len(isInt), isInt
print ' isFloat:', len(isFloat), isFloat
print 'isString:', len(isString), isString
print ' other:', len(other), other
# my output and the output I wanted
'''
MIXED_DATA_TYPES total count: 14
isClass: 2 ['Class_1', 'Class_2']
isList: 2 ['list_1', 'list_2']
isTuple: 2 ['tuple_1', 'tuple_2']
isDict: 2 ['dict_1', 'dict_2']
isInt: 2 ['x', 'y']
isFloat: 2 ['pie', 'eee']
isString: 2 ['house', 'cabin']
other: 0 []
'''
you can use easydict
>>> from easydict import EasyDict as edict
>>> d = edict({'foo':3, 'bar':{'x':1, 'y':2}})
>>> d.foo
3
>>> d.bar.x
1
>>> d = edict(foo=3)
>>> d.foo
3
another example:
>>> d = EasyDict(log=False)
>>> d.debug = True
>>> d.items()
[('debug', True), ('log', False)]
On python3, this function will get the outer most name in the stack:
import inspect
def retrieve_name(var):
"""
Gets the name of var. Does it from the out most frame inner-wards.
:param var: variable to get name from.
:return: string
"""
for fi in reversed(inspect.stack()):
names = [var_name for var_name, var_val in fi.frame.f_locals.items() if var_val is var]
if len(names) > 0:
return names[0]
It is useful anywhere on the code. Traverses the reversed stack looking for the first match.
While this is probably an awful idea, it is along the same lines as rlotun's answer but it'll return the correct result more often.
import inspect
def getVarName(getvar):
frame = inspect.currentframe()
callerLocals = frame.f_back.f_locals
for k, v in list(callerLocals.items()):
if v is getvar():
callerLocals.pop(k)
try:
getvar()
callerLocals[k] = v
except NameError:
callerLocals[k] = v
del frame
return k
del frame
You call it like this:
bar = True
foo = False
bean = False
fooName = getVarName(lambda: foo)
print(fooName) # prints "foo"
should get list then return
def get_var_name(**kwargs):
"""get variable name
get_var_name(var = var)
Returns:
[str] -- var name
"""
return list(kwargs.keys())[0]
It will not return the name of variable but you can create dictionary from global variable easily.
class CustomDict(dict):
def __add__(self, other):
return CustomDict({**self, **other})
class GlobalBase(type):
def __getattr__(cls, key):
return CustomDict({key: globals()[key]})
def __getitem__(cls, keys):
return CustomDict({key: globals()[key] for key in keys})
class G(metaclass=GlobalBase):
pass
x, y, z = 0, 1, 2
print('method 1:', G['x', 'y', 'z']) # Outcome: method 1: {'x': 0, 'y': 1, 'z': 2}
print('method 2:', G.x + G.y + G.z) # Outcome: method 2: {'x': 0, 'y': 1, 'z': 2}
With python-varname you can easily do it:
pip install python-varname
from varname import Wrapper
foo = Wrapper(True)
bar = Wrapper(False)
your_dict = {val.name: val.value for val in (foo, bar)}
print(your_dict)
# {'foo': True, 'bar': False}
Disclaimer: I'm the author of that python-varname library.
>>> a = 1
>>> b = 1
>>> id(a)
34120408
>>> id(b)
34120408
>>> a is b
True
>>> id(a) == id(b)
True
this way get varname for a maybe 'a' or 'b'.
if i declare an enum like
enum Weekdays
{
Mon = 1,
Tue = 1,
Wen = 1,
Thi,
Fri,
Sat,
Sun
}
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);//Prints Tue why?
now if i change Weekdays and do the same operation as follows
enum Weekdays
{
Mon = 1,
Tue = 1,
Wen = 1,
Thi = 1,
Fri,
Sat,
Sun
}
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);//Prints Thi !!!!!How?
What is really happening here?
When you write out your value like this, it ends up getting ToString() called on it.
Console.WriteLine(obj);
If you dig into the code far enough down, it's calling Enum.GetName() on your value.
Regarding multiple enum values with the same underlying value, the Enum.GetName page on MSDN says:
If multiple enumeration members have the same underlying value, the GetName method guarantees that it will return the name of one of those enumeration members. However, it does not guarantee that it will always return the name of the same enumeration member. As a result, when multiple enumeration members have the same value, your application code should never depend on the method returning a particular member's name.
It doesn't state how it determines which name to return if the values on two or more are the same.
The docs for Enum.ToString() include the same warning, in slightly different wording.
Digging a little deeper, the method above makes a call to Array.BinarySearch, passing it an array of numbers representing all values in your enum, and a number representing the value you want to print.
So you have an array with multiple 1's in it, and you're searching for a 1. The docs for that call are similar:
Duplicate elements are allowed. If the Array contains more than one element equal to value, the method returns the index of only one of the occurrences, and not necessarily the first one.
Again, it doesn't state how a selection is made, just that it'll be unreliable.
When you assign similar values, the result will be unexpected but I think it will evaluate for two cases:
When n is even:
(n/2)
When n is odd:
(n/2)+1
If I change the enum like this:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1,Fri=1,Sat=1, Sun=1, Mon2=1, Mon3=1}
// n is odd = 9
// (n/2)+1 = 5
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result will be Fri, Now lets change the enum again:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1,Fri=1,Sat=1, Sun=1,Mon2=1}
// n is even = 8
// (n/2) = 4
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Thi, Again change the enum:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1,Fri=1,Sat=1, Sun=1}
// n is odd = 7
// (n/2)+1 = 4
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Thi, Again change the enum:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1,Fri=1,Sat=1}
// n is even = 6
// (n/2) = 3
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Wen, Again change the enum:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1,Fri=1}
// n is odd = 5
// (n/2)+1 = 3
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Wen, Changing the enum again:
enum Weekdays {Mon=1,Tue=1,Wen=1,Thi=1}
// n is even = 4
// (n/2) = 2
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Tue, Changing the enum again:
enum Weekdays {Mon=1,Tue=1,Wen=1}
// n is odd = 3
// (n/2)+1 = 2
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);
The result is now Tue.
Even though this is explaining the behavior perfectly but this may not always happen or may not happen since I have not checked this for more cases but as MSDN says you should not assume about such output when the enum have same values for different names...
That said, I think you can easily understand now what is happening in your code.
Ref.: Link
Edit:
#GrantWinney's answer led me to this, He has written that Array.BinarySearch is passed the array of values and the value to search for so I realized from the name Array.BinarySearch that it is definitely using a BinarySearch and that explains everything...
Binary Search will divide the array like this:
Mid = {Low(which is the starting index) + High (which is the last index of array)}/2
and then Check for
if (Mid == value) return index;
else
if the value is smaller or equal move left other wise move right of the array
So this explains it how the enum values are printed if their are multiple names for the value you are trying to print.
Your Original Question
enum Weekdays
{
Mon = 1,
Tue = 1,
Wen = 1,
Thi,
Fri,
Sat,
Sun
}
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);//Prints Tue why?
It prints Tue because a call to Array.BinarySearch will be made passing the array
{1, 1, 1, 2, 3, 4, 5}
and a value to search which is 1...
So the BinarySearch will do this:
Mid = {Low(0) + High(6)} / 2
if (Mid == value) return index
else move left
After moving left again the Mid will be calculated:
High = Mid - 1; // now only the left sub-array will be searched
Mid = {Low(0) + High(2)} / 2
if (Mid == value) return index // here the condition will be true and you will be returned with `Tue`
The 2nd example in your Question:
enum Weekdays
{
Mon = 1,
Tue = 1,
Wen = 1,
Thi = 1,
Fri,
Sat,
Sun
}
Weekdays obj = (Weekdays)1;
Console.WriteLine(obj);//Prints Thi !!!!!How?
As I have written above a call to Array.BinarySearch will be made and array:
{1, 1, 1, 1, 2, 3, 4}
will be passed with value = 1 to search...
Apply the BinarySearch algorithm on the array and it will evaluate to Thi.
As Per MSDN If multiple enumeration members have the same underlying value and you attempt to retrieve the string representation of an enumeration member's name based on its underlying value, your code should not make any assumptions about which name the method will return.
I have a Vector class, and I was testing the following unit test (using nUnit).
1 Vector test1 = new Vector(new double[] { 6, 3, 4, 5 });
2 Vector test2 = test1;
3 Assert.AreEqual(test1, test2, "Reference test");
4 test2 = new Vector(new double[] { 3, 3, 4, 5 });
5 Assert.AreEqual(test1, test2, "Reference test");
The first test in line 3 passes, but the second test in line 5 fails. Shouldn't test2 also point to the same memory as test1, since I did the assignment statement in line 2? My Vector is defined as a class, so it is a reference type. On the other hand, the following tests pass:
1 Vector test1 = new Vector(new double[] { 6, 3, 4, 5 });
2 Vector test2 = test1;
3 Assert.AreEqual(test1, test2, "Reference test");
4 test2[1] = 4;
5 Assert.AreEqual(test1, test2, "Reference test");
Does that mean that, when I use the new operator to define a new object, old assignments are no longer valid? Any other (or correct - if I am wrong) explanation?
The line
test2 = new Vector(new double[] { 3, 3, 4, 5 });
creates a new instance of Vector on the heap and assigns its address to the test2 variable. test2 will point to a new, completely distinct object after that.
In contrast, the line
test2[1] = 4;
does not change the test2 variable itself (which is a reference to some object on the heap). Rather, it's changing the object it points to. test2 still refers to the same location on the heap.
To summarize, in the former, you are changing the reference while in the latter, you are altering the referent.
When you assign a variable like:
test2 = new Vector(new double[] { 3, 3, 4, 5 });
You are changing the value of test2 to be the new reference returned by the right hand side of the assignment operator. Of course the reference returned here is different than the one in test1, because it is a separate case of the constructor being called, and the references could not possibly be the same, as the Vector is being constructed with different arguments.
Yes, when you use the new operator to define a new object, old assignments are no longer valid.
Your vector IS a reference type, but when you say test2 = something you are saying "now test2 points to something else".
As an aside, if you want two different Vector objects with the same internal values to be considered equal, you can get that by implementing IEquatable on your Vector class, but that's another question...
Equals compares the values to see if they match, whereas if you want to compare references you need to use ReferenceEquals.
Check out http://msdn.microsoft.com/en-us/library/dd183752.aspx
Vector test1 = new Vector(new double[] { 6, 3, 4, 5 });
Is creating two objects, the vector its self, and the reference to it.
Imagin you have a list of items on a page.
When you use new Vector you effectively write a new line on the page which contains the vector.
Objects
{1,2,3,4,5}
You also have a list of references (Vector test1 = new Vector) which referance the first page, and (test2 = test1)
References
test1-> 1
test2-> 1
when you say 'test2 = new Vector {5,4,3,2,1} you then end up with a new vector object on the first page, and change which vector test2 is referring to.
Objects
{1,2,3,4,5}
{5,4,3,2,1}
References
test1 -> 1
test2 -> 2
In your second example both test1 and test2 are still pointing to the same object, which is why the test passes.