Following comparison in SageMath always gives True, meaning, in the first if condition it prints out 'Not zero', and in the second one it prints out 'Zero'.
a = sin(1/2*arctan2(0, sqrt(3)-1))
if 0 != a:
print('Not zero')
else:
print('Zero')
if 0 == a:
print('Zero')
else:
print('Not zero')
Is this supposed to be undefined behaviour or am I missing something?
I know that using .n(num_dec) on a will give the correct answer, but I'm interested in why this comparison always returns True.
Related
Hey so I am having an issue with an OR and ternary operation. I have this code here.
console.log(this.state.records)
Before: {
AllowCheck: "1"
Checked: "0"
}
const records = this.state.records.map((record) => {
return {
...record,
Checked: checked || record.AllowCheck === '0' ? '1' : '0',
}
});
console.log(records)
After: {
AllowCheck: "1"
Checked: "1"
}
When you look at the log before the ternary operation, you see that the object has a property called “AllowCheck”. You can see here that it evaluates as a 1 in the record. If you look at the function below, you’ll see a map operation that iterates over a list of records. The variable “checked” comes from a checkbox onChange operation that will evaluate as true in this situation. In the OR operation you can see that “checked” will be true, and the ternary on the right is where the “record.AllowCheck” will evaluated as a 1 from before. The ternary should result in a 0 since “record.AllowCheck” is 1. You’ll see in the after object that Checked is equal to 1. I don't know why it's not equal to two from the "checked" variables, and I really don't understand how it's equal to 1. Am I missing something? Have I been looking at this for too long? Any opinions or answers would be much appreciated, thank you.
The ternary should result in a 0 since “record.AllowCheck” is 1
Nope. The condition in this ternary operation is not what you think it is (record.AllowCheck === '0'). It's actually checked || record.AllowCheck === '0' and, since checked is truthy, it short-circuits on the first step, evaluates to true overall and that's how the ternary operator evaluates to '1'.
See the operator precedence table for more information: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence#table
I'm trying to determine if a value is found in the binary search tree.
If it's found, the value is printed. If not, a message is printed saying it wasn't found.
My problem is that even when the value is found, the message is printed saying that it wasn't found.
My result seems to reset even after returning True, and I am confused as to why this is happening... I think it's because I'm calling the function recursively, but I don't know how to fix this problem. Any help would be appreciated, thank you.
def lookUpVal(bst,val,result):
if bst == None:#base case, if the tree is 0, return none
return
elif bst['data'] == val:
print ("value found")
result = True
return result
lookUpVal(bst['left'],val,result)
lookUpVal(bst['right'],val,result)
def main(bst):
print ("Enter the value you want to find")
val = int(input())
result = 0
lookUpVal(bst,stud,result)
if result != True:
print ("Value not found")
The problem is with your result variable, you probably think you are passing by reference, what is actually happening is close to pass-by-value.
Here is an example:
def voo(x):
print('x:',id(x))
x = True
print('x:',id(x))
p = False
print('p:',id(p))
voo(p)
print('value of p:',p)
print('p:',id(p))
'id' returns a unique id for any object, including of course boolean ones.
Here's the output: (numbers will vary in your pc)
p: 1613433952
x: 1613433952
x: 1613433936
value of p: False
p: 1613433952
First note, False in output, p's value has not changed. But to see why that's happening, closely examine the id values, specially how x's id changed after assignment in function; which means python allocated a new object. And old one is still referenced by 'p', as evident in it's output, which has not changed.
I'm working my way through the Data Science courses at DataCamp. (Not a plug.) One of the practice lessons has the following completed solution:
# logs is available in your workspace
extract_info <- function(x, property = "success", include_all = TRUE) {
info <- c()
for (log in x) {
if (include_all || !log$success) {
info <- c(info, log[[property]])
}
}
return(info)
}
# Call extract_info() on logs, no additional arguments
extract_info(logs)
# Call extract_info() on logs, set include_all to FALSE
extract_info(logs, include_all = FALSE)
The first call (extract_info(logs)) works as I would expect it to: it returns a vector containing all the log entries (regardless of the value of log$success).
The second call (extract_info(logs, include_all = FALSE)) does not return the results I would expect. It returns a vector containing only those results where log$success evaluates to FALSE.
It seems to me that the use of the || operator should cause the if block to short-circuit, and the second call should return nothing. From what I can tell, R evaluates expressions from left to right; but this looks like it's evaluating from right to left.
Can someone explain what's going on here?
(According to the site, this is the correct solution, and there's nothing wrong with the code. I want to know why it works the way it does. Even if the answer is that I'm overlooking something painfully obvious and stupid.)
Well || is the "or" operator. You don't short circuit the "or" operator with a FALSE value; you basically ignore that parameter and just look at the next one because you are looking for any TRUE value.
Assume a is a boolean value. These should be equivalent (<==>).
# or
FALSE || a <==> a
TRUE || a <==> TRUE
# and
TRUE && a <==> a
FALSE && a <==> FALSE
It seems like this was a temporary confusion.
|| is OR and so if either condition evaluates to TRUE, the compound expression evaluates to TRUE. If include_all was TRUE, you could short-circuit the expression, but when include_all is FALSE, you must wait to see what the other part is.
Simple question. I tried searching, by googling for less than and greater than signs doesn't return great results.
My guess is that <> is basically equivalent to not equals. So, the below expression would be false if x is null or an empty string, and true otherwise?
if x <> ""
This would also return True if a value is contained in the entity listed. This is commonly used to look for quesrystring or form elements that may or may not have been supplied:
If Request("someFieldName") <> "" Then
' Field was provided and has a value, so use the field value
Else
' Field was either empty or not provided, in which case use something else
End If
Hope this helps.
So, the below expression would be false if x is null or an empty string, and true otherwise?
Not exactly. There are few function to verify value:
IsNull(expression)
IsNull returns True if expression is Null, that is, it contains no
valid data; otherwise, IsNull returns False. If expression consists of
more than one variable, Null in any constituent variable causes True
to be returned for the entire expression.
The Null value indicates that the variable contains no valid data.
Null is not the same as Empty, which indicates that a variable has not
yet been initialized. It is also not the same as a zero-length string
(""), which is sometimes referred to as a null string.
IsEmpty(expression)
The expression argument can be any expression. However, because
IsEmpty is used to determine if individual variables are initialized,
the expression argument is most often a single variable name.
IsEmpty returns True if the variable is uninitialized, or is
explicitly set to Empty; otherwise, it returns False. False is always
returned if expression contains more than one variable.
Other good function
VarType(varname)
Returns a value indicating the subtype of a variable.
Use Windows Script 5.6 Documentation from http://www.microsoft.com/en-us/download/details.aspx?id=2764
This example can be easily tested in the groovy console.
var a is evaluated to not null while b is evaluated to null.
Both are instances of org.codehaus.groovy.runtim.NullObject
def b = null
println b.getClass()
println b == null
def a = null.getClass().newInstance()
println a.getClass()
println a == null
Does anyone knows why?
This is a tricky thing when dealing with reflection code.
Actually I am wondering if this is not a bug. As an explanation... NullObject is a runtime/intermediate kind of Object. If you do anything on null, then NullObject is used. This, and the the implementation of NullObject#equals speaks for a==null returning true. It returns fails, because there is some internal code before that, that is for example determining if compareTo is called instead of equals and such things. Now this piece of code starts with
if (left == right) return true;
if (left == null || right == null) return false;
so null==null will return true, but NullObject==null will return false. On the other hand NullObject should not leak out if possible. Maybe we should fix newInstance() to return null.
I filled http://jira.codehaus.org/browse/GROOVY-5769 for this
In the equals method of NullObject, it only returns true if you are comparing it to null
As an instance of NullObject is not strictly null, it returns false...
Whether NullObject should return true if you call equals against another NullObject is probably a question best asked on the mailing list... I'll have a look and see if I can find any previous question.