I'm getting incorrect Mod operator result in VB.NET. I verified result with calculator, It gives correct result.
E.g In VB.NET 1.3 Mod 0.05 = 0.049999999999999975 whereas in Calculator it shows 0
It's round up error:
1.30000000000001 Mod 0.05 == 0.00000000000001
1.2999999 Mod 0.05 == 0.0499998999999999
usually, round up errors are little (if any) nuisance, put at / near points of discontinuity small errors lead to big difference at the result (0.05 in this case).
Amendment: change the initial double to either int (long)
130 Mod 5 == 0
or decimal:
1.3M Mod 0.05M == 0
Related
I am currently building my first stuff on Smalltalk and I have hit an issue. I have to deal with a user-entered number, and I need to div it by 2 and still be an integer. If an user inputs 10, I will work with 5, if they input 11, I have to work with 6, but I will obviously get 5.5.
If I could get the mod of a number I could simply make sure mod = 0 else add 0.5 and it would do just as good, but I just can't find how to make a mod operation in SmallTalk, all my searches end up in unrelated stuff about actual social smalltalk, which is extremely frustrating.
So if you could tell me how to get the mod of a number it would be great, if you could tell me how to round up with a separate function, even better. Thanks for your help and time beforehand.
UPDATE: After some research, I tried to do it this way:
mod := par rem: 2.
mod = 0 ifFalse: [ par := par + 0.5 ].
where as "mod" is mod of the variable "par", and if it isn't 0, it should add up 0.5 to par.
My issue now is that trying to use par in a timesRepeat brings up a "BoxedFloat64 did not understand #timesRepeat" error. So I am still in the same issue, or just need a way to make a float into an integer.
There are a lot of ways. For example
Add 1 to entered number before div by 2 if entered number is odd
temp := enteredNumber.
temp odd ifTrue: [temp := temp + 1 ].
^temp / 2
Using ceiling method
^(enteredNumber / 2) ceiling
In Smalltalk, we have an operator for integer division (and even two operators):
11 / 2
would answer a Fraction, not a whole Integer
But:
11 // 2
would answer the quotient of division, rounded toward negative infinity
And the corresponding remainder will be:
11 \\ 2
The second operator quo: for quotient and rem: for remainder
The difference is only with negative receiver/operand: the later ones are truncating the quotient toward zero.
-11 // 4 = -3. "floored toward negative infinity"
-11 \\ 4 = 1.
(-11 quo: 4) = -2. "truncated toward zero"
(-11 rem: 4) = -3.
If you want to round the quotient upper (toward positive infinity), then you can write:
(anInteger + 1) // 2.
Or same without parenthesis if you are confident enough in binary operator precedence:
anInteger + 1 // 2.
I have this code I need to explain and the Mod operator has confused me, MSDN has a page on it and I cannot understand it clearly. I have included a section of code below if you could refer to it in your answer, thanks.
number1 = (input1/ 10) - 0.5
number2 = input2 Mod 10
Result = number1 + number2
Mod in VB.NET is the Modulo operation. It returns the remainder when one number is divided by another.
For example, if you divided 4 by 2, your mod result would be 0 (no remainder). If you divided 5 by 2, your mod result would be 1.
Explanation of the Modulo operation
I need to divide one int into 2 other int's. the first int is not constant so one problem would be, what to do with odd numbers because I only want whole numbers. For example, if int = 5, then int(2) will = 2 and int(3) will = 3. Any help will greatly be appreciated.
Supposing you want to express x = a + b, where a and b are as close to x/2 as possible:
a = ceiling(x / 2.0);
b = floor(x / 2.0);
That's pseudo code, you have to find out the actual functions for floor and ceiling from your library. Make sure the division is performed as floating point numbers.
As pure integers:
a = x / 2 + (x % 2 == 0 ? 0 : 1);
b = x / 2
(This may be a bit fishy for negative numbers, because it'll depend on the behaviour of division and modulo for negative numbers.)
You can try ceil and floor functions from math to produce results like 2 and 3 for odd inputs;
int(2)=ceil(int/2); //will produce 3 for input 5
int(3)=floor(int/2); //will produce 2 for input 5
Well my answer is not in Objective-C but i guess you could translate this easily.
My idea is:
part1 = source_number div 2
part2 = source_number div 2 + (source_number mod 2)
This way the second number will be bigger if the starting number is an odd number.
I'm trying to make a generic equation which converts a value. Here are some examples.
9,873,912 -> 9,900,000
125,930 -> 126,000
2,345 -> 2,400
280 -> 300
28 -> 30
In general, x -> n
Basically, I'm making a graph and I want to make values look nicer. If it's a 6 digit number or higher, there should be at least 3 zeros. If it's a 4 digit number or less, there should be at least 2 digit numbers, except if it's a 2 digit number, 1 zero is fine.
(Ignore the commas. They are just there to help read the examples). Anyways, I want to convert a value x to this new value n. What is an equation g(x) which spits out n?
It is for an objective-c program (iPhone app).
Divide, truncate and multiply.
10**x * int(n / 10**(x-d))
What is "x"? In your examples it's about int(log10(n))-1.
What is "d"? That's the number of significant digits. 2 or 3.
Ahhh rounding is a bit awkward in programming in general. What I would suggest is dividing by the power of ten, int cast and multiplying back. Not remarkably efficient but it will work. There may be a library that can do this in Objective-C but that I do not know.
if ( x is > 99999 ) {
x = ((int)x / 1000) * 1000;
}
else if ( x > 999 ) {
x = ((int) x / 100) * 100;
}
else if ( x > 9 ) {
x = ((int) x / 10) * 10;
}
Use standard C functions like round() or roundf()... try man round at a command line, there are several different options depending on the data type. You'll probably want to scale the values first by dividing by an appropriate number and then multiplying the result by the same number, something like:
int roundedValue = round(someNumber/scalingFactor) * scalingFactor;
So I thought that negative numbers, when mod'ed should be put into positive space... I cant get this to happen in objective-c
I expect this:
-1 % 3 = 2
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
But get this
-1 % 3 = -1
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
Why is this and is there a workaround?
result = n % 3;
if( result < 0 ) result += 3;
Don't perform extra mod operations as suggested in the other answers. They are very expensive and unnecessary.
In C and Objective-C, the division and modulus operators perform truncation towards zero. a / b is floor(a / b) if a / b > 0, otherwise it is ceiling(a / b) if a / b < 0. It is always the case that a == (a / b) * b + (a % b), unless of course b is 0. As a consequence, positive % positive == positive, positive % negative == positive, negative % positive == negative, and negative % negative == negative (you can work out the logic for all 4 cases, although it's a little tricky).
If n has a limited range, then you can get the result you want simply by adding a known constant multiple of 3 that is greater that the absolute value of the minimum.
For example, if n is limited to -1000..2000, then you can use the expression:
result = (n+1002) % 3;
Make sure the maximum plus your constant will not overflow when summed.
We have a problem of language:
math-er-says: i take this number plus that number mod other-number
code-er-hears: I add two numbers and then devide the result by other-number
code-er-says: what about negative numbers?
math-er-says: WHAT? fields mod other-number don't have a concept of negative numbers?
code-er-says: field what? ...
the math person in this conversations is talking about doing math in a circular number line. If you subtract off the bottom you wrap around to the top.
the code person is talking about an operator that calculates remainder.
In this case you want the mathematician's mod operator and have the remainder function at your disposal. you can convert the remainder operator into the mathematician's mod operator by checking to see if you fell of the bottom each time you do subtraction.
If this will be the behavior, and you know that it will be, then for m % n = r, just use r = n + r. If you're unsure of what will happen here, use then r = r % n.
Edit: To sum up, use r = ( n + ( m % n ) ) % n
I would have expected a positive number, as well, but I found this, from ISO/IEC 14882:2003 : Programming languages -- C++, 5.6.4 (found in the Wikipedia article on the modulus operation):
The binary % operator yields the remainder from the division of the first expression by the second. .... If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined
JavaScript does this, too. I've been caught by it a couple times. Think of it as a reflection around zero rather than a continuation.
Why: because that is the way the mod operator is specified in the C-standard (Remember that Objective-C is an extension of C). It confuses most people I know (like me) because it is surprising and you have to remember it.
As to a workaround: I would use uncleo's.
UncleO's answer is probably more robust, but if you want to do it on a single line, and you're certain the negative value will not be more negative than a single iteration of the mod (for example if you're only ever subtracting at most the mod value at any time) you can simplify it to a single expression:
int result = (n + 3) % 3;
Since you're doing the mod anyway, adding 3 to the initial value has no effect unless n is negative (but not less than -3) in which case it causes result to be the expected positive modulus.
There are two choices for the remainder, and the sign depends on the language. ANSI C chooses the sign of the dividend. I would suspect this is why you see Objective-C doing so also. See the wikipedia entry as well.
Not only java script, almost all the languages shows the wrong answer'
what coneybeare said is correct, when we have mode'd we have to get remainder
Remainder is nothing but which remains after division and it should be a positive integer....
If you check the number line you can understand that
I also face the same issue in VB and and it made me to forcefully add extra check like
if the result is a negative we have to add the divisor to the result
Instead of a%b
Use: a-b*floor((float)a/(float)b)
You're expecting remainder and are using modulo. In math they are the same thing, in C they are different. GNU-C has Rem() and Mod(), objective-c only has mod() so you will have to use the code above to simulate rem function (which is the same as mod in the math world, but not in the programming world [for most languages at least])
Also note you could define an easy to use macro for this.
#define rem(a,b) ((int)(a-b*floor((float)a/(float)b)))
Then you could just use rem(-1,3) in your code and it should work fine.