what formula can identify an ODD vs EVEN number
ODD result can be 0
EVEN result can be 1
or vice versa
The expression:
Mod ( YourNumber ; 2 )
returns 1 (True) if YourNumber is odd, 0 (False) otherwise.
The expression:
not Mod ( YourNumber ; 2 )
returns 1 (True) if YourNumber is even, 0 (False) otherwise.
Related
I understand the whole code and
I just want to know why there has to be a -1 at the end of the range function.
I've been checking it out with pythontutor but I can't make it out.
#Given 2 strings, a and b, return the number of the positions where they
#contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3,
#since the "xx", "aa", and "az" substrings appear in the same place in
#both strings.
def string_match(a, b):
shorter = min(len(a), len(b))
count = 0
for i in range(shorter -1): #<<<<<<<<< This is -1 I don't understand.
a_sub = a[i:i+2]
b_sub = b[i:i+2]
if a_sub == b_sub:
count = count + 1
return count
string_match('xxcaazz', 'xxbaaz')
string_match('abc', 'abc')
string_match('abc', 'axc')
I expect to understand why there has to be a -1 at the end of the range function. I will appreciate your help and explanation!
The value indices of the for loop are counted since 0 so the final value actually would be the (size -1)
I have the following 3 examples of case expressions in postgres, which I would expect to evaluate in the same way. However the first and the third give ERROR: invalid input syntax for integer: "2017,7". The second one seems to be ok. Why is the difference?
Postgres documentation states
"A CASE expression does not evaluate any subexpressions that are not
needed to determine the result."
select case when 0 = 0 then 1 < 2
when 0 = 2 then ('2017.7')::bigint > 2000
end
(DB Fiddle)
select case when 0 = 0 then 1 < 2
when 0 = 2 then 2000 = ('2017.7'||'')::bigint
end
(DB Fiddle)
select case when (array[1,2])[1] =1 then 1 < 2
when (array[1,2])[1] = 2 then 2000 = ('2017.7'||'')::bigint
end
(DB Fiddle)
if ((status & 0x3F) == 1 ){ }..
the status is variable in swift language.
what is mean about this condition, & mean and (status & 0x3F) value return
& is the bitwise AND operator. It compares the bits of the two operands and sets the corresponding bit to 1 if it is 1 in both operands, or to 0 if either or both are 0.
So this statement:
((status & 0x3F) == 1)
is combining status with 0b111111 (the binary equivalent of 0x3F and checking if the result is exactly 1. This will only be true if the last 6 bits of status are 0b000001.
In this if:
if( (dtc24_state[2] & 0x8) == 0x8 ) {
self.haldexABCDTC24State.text = status_str + " - UNKNOWN"
self.haldexABCDTC24State.textColor = text_color
active_or_stored_dtc = true
}
dct24_state is an array of values. The value of dct24_state[2] is combined with 0x8 or 0b1000 and checked against 0x8. This is checking if the 4th bit from the right is set. Nothing else matters. If the 4th bit from the right is set, the if is true and the code block is executed.
0x3F is 111111. So, it means this:
for each bit of yourNumber in binary system presentation use and method.
This way truncates the left part of the number. and the result compares with 1.
e.g.
7777 is 1111001100001 after executing and this number converts into
100001. So the result is false.
But for 7745 (1111001000001) the result is 1. The result is true.
The rule for 'and' function: 0 & 0 = 0 ; 0 & 1 = 0; 1 & 0 = 1; 1 & 1 = 1.
7 -> 1
0 -> 0
-7 -> -1
I've have code:
(x == 0 ? 0 : x / abs(x)) + 1
but is it possible to avoid division and make it faster?
How about
(x == 0 ? 0 : (x < 0 ? -1 : 1))
The idea was to use bit operations to avoid branching code or value conversion.
Haven't found how to do it with bit operations but apple already add this function
https://developer.apple.com/documentation/swift/int/2886673-signum
signum()
Returns -1 if this value is negative and 1 if it’s positive; otherwise, 0.
so simple) raw test shows ~x100 faster implementation
I'm trying to create a decimal formatting formula on my summary fields. The values in the database could have 0, 1, or 2 decimal places. I've started with this:
If (CurrentFieldValue mod 1 = 0) Then
0
Else If (CurrentFieldValue mod .1 = 0) Then
1
Else
2
On a simple single data field, this works and displays the value with 0, 1, or 2 decimal places based on the data coming from my database. The same formula doesn't work for a summary field on my reports with group data. Any ideas?
Edit: Since I don't know how to format code in a comment, I'll address the suggestion of using a formula here:
Didn't work. Formula:
Sum ({myTable.dataValue}, {myTable.groupField})
then I used:
If ({#formula} mod 1 = 0) Then
0
Else If ({#formula} mod .1 = 0) Then
1
Else
2
And I still got whole numbers for everything. My rounding is set to .01 with no formula. Do I need a formula for rounding too? I still don't understand why this works on individual values but not for group summaries.
OK- it turns out this is due to our lack of understanding of the mod function :)
Everything mod 1 actually returns 0. This is the formula you need to use:
if {ER100_ACCT_ORDER.ER100_ORD_TOT} * 100 mod 100 = 0 then
0
else if {ER100_ACCT_ORDER.ER100_ORD_TOT} * 100 mod 10 = 0 then
1
else
2
:)
How about just creating a formula field instead of using the built-in summary field:
sum({mytable.myfield})
Then you can use your conditional formatting:
If ({#formula} mod 1 = 0) Then
0
Else If ({#formula} mod .1 = 0) Then
1
Else
2