is there a .NET library that can perform a numerical operation and return a value?
I have an expression like so:
1 + 1/2
this should return the double equivalent of the same.
I wont be passing a string, it will be a numeric value and the return should be a numeric.
I've used NCalc for medium-to-complex calculations and it seems that it will fit your needs.
About the OP's comment on the original question:
double d = 1/2;
will return 0 in c#, because the integer 1 is divided by the integer 2, resulting in an integer result of 0.
If you need to trigger real-number mathematics, at least one of the operands must be defined as a real number. You can do that by specifying it with a decimal point (1.0 instead of 1), or by adding a type specifier after the value (1f or 1d instead of 1).
Take a look at this example:
double d1 = (1 + 1/2); //returns 1
double d2 = (1 + 1.0/2); //returns 1.5
double d3 = (1 + 1/2.0); //returns 1.5
double d4 = (1 + 1f/2); //returns 1.5
double d5 = (1 + 1d/2); //returns 1.5
See here CodeDom Calculator - Evaluating C# Math Expressions Dynamically
Related
This simple calculation is returning zero, I can't figure it out:
decimal share = (18 / 58) * 100;
You are working with integers here. Try using decimals for all the numbers in your calculation.
decimal share = (18m / 58m) * 100m;
18 / 58 is an integer division, which results in 0.
If you want decimal division, you need to use decimal literals:
decimal share = (18m / 58m) * 100m;
Since some people are linking to this from pretty much any thread where the calculation result is a 0, I am adding this as a solution as not all the other answers apply to case scenarios.
The concept of needing to do calculations on various types in order to obtain that type as a result applies, however above only shows 'decimal' and uses it's short form such as 18m as one of the variables to be calculated.
// declare and define initial variables.
int x = 0;
int y = 100;
// set the value of 'x'
x = 44;
// Results in 0 as the whole number 44 over the whole number 100 is a
// fraction less than 1, and thus is 0.
Console.WriteLine( (x / y).ToString() );
// Results in 0 as the whole number 44 over the whole number 100 is a
// fraction less than 1, and thus is 0. The conversion to double happens
// after the calculation has been completed, so technically this results
// in 0.0
Console.WriteLine( ((double)(x / y)).ToString() );
// Results in 0.44 as the variables are cast prior to calculating
// into double which allows for fractions less than 1.
Console.WriteLine( ((double)x / (double)y).ToString() );
Because the numbers are integers and you perform integer division.
18 / 58 is 0 in integer division.
Whenever I encounter such situations, I just upcast the numerator.
double x = 12.0 / 23409;
decimal y = 12m / 24309;
Console.WriteLine($"x = {x} y = {y}");
double res= (firstIntVar * 100f / secondIntVar) / 100f;
when dividing numbers I use double or decimal , else I am getting 0 , with this code even if firstIntVar && secondIntVar are int it will return the expected answer
decimal share = (18 * 100)/58;
Solved: working perfectly with me
int a = 375;
int b = 699;
decimal ab = (decimal)a / b * 100;
Here is the code which made me post this question.
// int integer;
// int fraction;
// double arg = 110.1;
this.integer = (int)(arg);
this.fraction = (int)((arg - this.integer) * 100);
The variable integer is getting 110. That's OK.
The variable fraction is getting 9, however I am expecting 10.
What is wrong?
Update
It seems I have discovered that the source of the problem is subtraction
arg - this.integer
Its result is 0.099999999999994316.
Now I am wondering how I should correctly subtract so that the result was 0.1.
You have this:
fraction = (int)((110.1 - 110) * 100);
The inner part ((110.1 - 110) * 100), will be 9.999999
When you cast it to int, it will be round off to 9
This is because of "floating point" (see here) limitations:
Computers always need some way of representing data, and ultimately
those representations will always boil down to binary (0s and 1s).
Integers are easy to represent, but non-integers are a bit more
tricky. Consider the following var:
double x = 0.1d;
The variable x will actually store the closest available double to
that value. When you understand this, it becomes obvious why some
calculations seem to be "wrong".
If you were asked to add a third to a third, but could only use 3
decimal places, you'd get the "wrong" answer: the closest you could
get to a third is 0.333, and adding two of those together gives 0.666,
rather than 0.667 (which is closer to the exact value of two thirds).
Update:
In financial applications or where the numbers are so important to be exact, you can use decimal data type:
(int)((110.1m - 110) * 100) //will be 10 (m is decimal symbol)
or:
decimal arg = 110.1m;
int integer = (int)(arg); //110
decimal fraction = (int)((arg - integer) * 100); //will be 10
It is because you are using double, precision gets rounded, if you want it to be 10 use decimal type:
check the following:
int integer;
int fraction;
decimal arg = 110.1M;
integer = (int)(arg);
decimal diff = arg - integer;
decimal multiply = diff * 100;
fraction = (int)multiply;//output will be 10 as you expect
I'm trying to do a simple picee of maths where I work out if a value is between two values and if so, it does, it should do a simple division. However sometimes the value divided by are like 0.9, 0.6 etc and that always returns 0.
in this example,
int m_LocationSqrMtr = 4339;
float m_DefaultPricing = Convert.ToSingle(DefaultPricing);
float m_manDays;
if (m_LocationCosts > 450 && m_LocationCosts < 700)
{
m_DefaultPricing = 700 / m_LocationSqrMtr;
}
My guess is that the type of m_LocationSqrMtr is int, in which case this expression:
700 / m_LocationSqrMtr
... will be computed using integer arithmetic, and the result converted to float. I suspect you want:
if (m_LocationCosts > 450 && m_LocationCosts < 700)
{
m_DefaultPricing = 700f / m_LocationSqrMtr;
}
The f suffix on the literal means that it's a float literal, so first m_LocationSqrMtr will be promoted to float, and then the division performed using float arithmetic.
However, if this is meant to be representing currency values, you should consider using decimal instead of float - and then probably rounding the value to 2 decimal places. If you do all your currency arithmetic in decimal, you're less likely to run into unexpected results...
You have:
int m_LocationSqrMtr = 4339;
[...]
m_DefaultPricing = 700 / m_LocationSqrMtr;
That is, 700 / 4339, which is (integer) / (integer), the result of which is an integer.
I know you were expecting an answer of 0.16132....
But in integer terms, that value is ZERO.
If the type of m_LocationSqrMtr is int, as in a 32-bit whole number, then the expression
700 / m_LocationSqrMtr
is one integer divided by another integer, and it's type is integer. Only after this integer is produced is the result assigned to the float m_DefaultPricing, so basically your code is equivalent to:
int temp = 700 / m_LocationSqrMtr;
m_DefaultPricing = (float) temp;
If you want to force floating point arithmetic, at least one of the operands needs to be a floating point number. There are a number of ways that this can be done:
m_DefaultPricing = 700.0 / m_LocationSqrMtr; //explicit decimal point
m_DefaultPricing = 700f / m_LocationSqrMtr; //explicit float specification
m_DefaultPricing = (float)700 / m_LocationSqrMtr; //casting one operand
m_DefaultPricing = 700f / (float)m_LocationSqrMtr; //into a float
There are already a number of good practical answers so on a more conceptual level:
In C# if both sides of the operator are an integer the result is also an integer and any decimal digits are truncated. In general, for math, the language will produce a result of the same type as the input with the most precise type. The type of variable you store the result in will have no effect on the result only the types of the inputs.
By forcing one input to be decimal or float you force the output to be that as well. In practice you can either do this by declaring your one of your input variables as a decimal or float (depending on which you use, in your case you would change the type of m_LocationSqrMtr) or if you have a constant input (as you do) you can force it to be decimal/float/double as follows:
var a = 10f // float
var b = 10d // double
var c = 10m // decimal
var d = 10.0 // double I believe
This question already has answers here:
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 9 years ago.
Today i come with a problem and not able to figure out what is the issue with this simple statement
I Tried
double d =1/4;
expected ans for me is 0.25 but in reality ans is 0.0 why so ??
And what should we do if statement is in terms of integer variables like this
double a =(a-b)/(d+e);
Because what you done is here integer division. 1 / 4 always give you 0 as a result regardless which type you assing it.
.NET has 3 type of division. From 7.7.2 Division operator
Integer division
Floating-point division
Decimal division
From Integer division part;
The division rounds the result towards zero, and the absolute value of
the result is the largest possible integer that is less than the
absolute value of the quotient of the two operands.
If you want to 0.25 as a result, you should define one of your values as a floating point.
You can use one of these;
double d = 1d / 4d;
double d = 1d / 4;
double d = 1 / 4d;
And what should we do if statement is in terms of integer variables
like this
double a =(a-b)/(d+e);
I assume your a, b, d and e are integers, you should use one of these then;
double a = (double)(a-b) / (double)(d+e);
double a = (a-b) / (double)(d+e);
double a = (double)(a-b) / (d+e);
double d =1d/4;
should work.
If you don't specify the type of your numbers, it is treated as Integer. And integer 1/4 will be zero.
Use this:
double d = (double) 1 / 4;
/ Operator (msdn)
When you divide two integers, the result is always an integer. For
example, the result of 7 / 3 is 2. To determine the remainder of 7 /
3, use the remainder operator (%). To obtain a quotient as a rational
number or fraction, give the dividend or divisor type float or type
double. You can assign the type implicitly if you express the dividend
or divisor as a decimal by putting a digit to the right side of the
decimal point.
Try this:
double d = 1.0 / 4.0;
This simple calculation is returning zero, I can't figure it out:
decimal share = (18 / 58) * 100;
You are working with integers here. Try using decimals for all the numbers in your calculation.
decimal share = (18m / 58m) * 100m;
18 / 58 is an integer division, which results in 0.
If you want decimal division, you need to use decimal literals:
decimal share = (18m / 58m) * 100m;
Since some people are linking to this from pretty much any thread where the calculation result is a 0, I am adding this as a solution as not all the other answers apply to case scenarios.
The concept of needing to do calculations on various types in order to obtain that type as a result applies, however above only shows 'decimal' and uses it's short form such as 18m as one of the variables to be calculated.
// declare and define initial variables.
int x = 0;
int y = 100;
// set the value of 'x'
x = 44;
// Results in 0 as the whole number 44 over the whole number 100 is a
// fraction less than 1, and thus is 0.
Console.WriteLine( (x / y).ToString() );
// Results in 0 as the whole number 44 over the whole number 100 is a
// fraction less than 1, and thus is 0. The conversion to double happens
// after the calculation has been completed, so technically this results
// in 0.0
Console.WriteLine( ((double)(x / y)).ToString() );
// Results in 0.44 as the variables are cast prior to calculating
// into double which allows for fractions less than 1.
Console.WriteLine( ((double)x / (double)y).ToString() );
Because the numbers are integers and you perform integer division.
18 / 58 is 0 in integer division.
Whenever I encounter such situations, I just upcast the numerator.
double x = 12.0 / 23409;
decimal y = 12m / 24309;
Console.WriteLine($"x = {x} y = {y}");
double res= (firstIntVar * 100f / secondIntVar) / 100f;
when dividing numbers I use double or decimal , else I am getting 0 , with this code even if firstIntVar && secondIntVar are int it will return the expected answer
decimal share = (18 * 100)/58;
Solved: working perfectly with me
int a = 375;
int b = 699;
decimal ab = (decimal)a / b * 100;