This question already has answers here:
Comparing double values in C#
(18 answers)
Closed 8 years ago.
I have a float value that i parsed into double later i rounded off to 2.Also i have another float value to which i did exactly the same as first one.
Here is the sample code..
string pulse = arrvaluedline[2].ToString();
pCost = float.Parse(arrvaluedline[3]);
double d = System.Convert.ToDouble(spCost);
double dd = Math.Round(d,2);
string[] arrpulse = pulse.Split(':');
vodanoofPulse = float.Parse(arrpulse[0]);
calculatedCost = CallCost * Pulse;
double dcalcost = Math.Round(calculatedCost, 2);
Now here i am trying to compare
if (dcalcost.Equals(spCost)){
}
Although my both values dcalcost and spCost are 0.4 Except this ,flow is not going inside the if ..Why..Please help me .,
The Equals method should be used with caution, because two apparently equivalent values can be unequal due to the differing precision of the two values. The following example reports that the Double value .333333 and the Double value returned by dividing 1 by 3 are unequal.
// Initialize two doubles with apparently identical values
double double1 = .33333;
double double2 = 1/3;
// Compare them for equality
Console.WriteLine(double1.Equals(double2)); // displays false
Comparing doubles are not as easy as one might think. Here is an example from MSDN on how you can do it in a better way.
// Initialize two doubles with apparently identical values
double double1 = .333333;
double double2 = (double) 1/3;
// Define the tolerance for variation in their values
double difference = Math.Abs(double1 * .00001);
if (Math.Abs(double1 - double2) <= difference)
Console.WriteLine("double1 and double2 are equal.");
else
Console.WriteLine("double1 and double2 are unequal.");
Related
This question already has answers here:
C# Maths gives wrong results!
(4 answers)
Closed 9 months ago.
Hi I have the function:
public static string MapDiePitchY(string DiePitchY)
{
float value = float.Parse(DiePitchY) * 1000;
int valInt = (int)value;
string temp = valInt.ToString();
return temp;
}
When I run it with these two numbers, I get two different behaviours:
str = MapDiePitchY("4150.8");
Returns 4150799
While
str = MapDiePitchY("2767.3");
Returns
2767300 which is what I'd expect and want everytime I pass in a float
I want the function to always return the value multiplied by 1000 but with no rounding. I've tried to replace the 1000 with 1000f and still get the behaviour where it adds extra value to 4150.8. Values that shouldn't exist there. I have no idea why this is happenign and googling hasn't given me any answers.
Is there a way to ensure I get the exact value I pass in as string but multiplied by 1000 with no rounding?
I am on C# 7.3
So you understand floating-point numbers are not exact. When you type float x = 4150.8f the internal bit representation of the number is
x = (1 + 112230 * 2^(-23)) * 2^(139-127) = 4150.7998046875
The integers m=112230 and e=139 represent the mantissa and exponent of the floating-point number.
The above is the result of the parse function. Then you multiply by 1000 which results in
value = x*1000 = 4150799.8046875
what you want to do at this point is do the rounding before converting into an integer
int valInt = (int)Math.Round(value);
which should round up the .80.. in the end into the next whole number 4150800.
This question already has answers here:
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 4 years ago.
I tried this:
var b = 100 / 95;
and b is declared as an integer. But I would like to have a number with a decimal point.
How can I do this in C#?
With two integer operands, the C# overload resolution picks a version of the division operator that does integer division.
To make C# choose real division, you need to change the type of at least one of the operands into decimal, float, or double.
You can achieve this by casting, in the general case. In your specific case, you can also just change the literals involved:
var b = 100m / 95; // decimal
var b = 100f / 95; // float
var b = 100.0 / 95; // double
Use decimal if these are exact, human-made quantities (e.g. money). Use float or double if you are dealing with approximate, physical quantities (e.g. irrational numbers, physical constants, etc).
You are doing an integer division which will ignore the fractional part of the division as it can not be stored in an int.
If you cast your variables to decimals you can store the fractional part of the division:
decimal b = (decimal)100 / (decimal)95;
This question already has answers here:
Why do these division equations result in zero?
(10 answers)
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 4 years ago.
I'm calculating the percentage to add in a datagrid, but when I report the data it always returns me 0, what am I doing wrong?
if I have the following case: the variable quantidadeInstalada has the value of 10 and the goal has 20 the concluido would have to return me 50 but it returns me 0
private void AdicionarPessoa()
{
string Valida = ValidaPessoa();
if (Valida.Equals(""))
{
double concluido=0, falta=0;
int quantidadeInstalada = Convert.ToInt32(ttbQuantidade.Text);
int meta = Convert.ToInt32(ttbMetaPessoa.Text);
concluido = (quantidadeInstalada/meta)*100;
falta = 100-concluido;
MessageBox.Show(concluido.Text);
}
else
MessageBox.Show(Valida);
}
It's probably due to the precision of the int. Use a decimal or double instead.
When we use an integer, we lose precision.
Console.WriteLine(100 / 17); // 5
Console.WriteLine(100 / 17m); // 5.8823529411764705882352941176
Console.WriteLine(100 / 17d); // 5.88235294117647
Console.WriteLine(100 / 17f); // 5.882353
Since integers always round down, 0.99 as an integer is 0.
Note that for precision, the types of the inputs matters.
double output = input1 * input2;
For example:
double outputA = 9 / 10;
Console.WriteLine(outputA); // 0
double outputB = 9 / 10d;
Console.WriteLine(outputB); // 0.9
double outputC = 9d / 10;
Console.WriteLine(outputC); // 0.9
Here is a Fiddle.
It's because you're doing integer division. You can fix it by casting one of the variables to a double like this:
concluido = ((double)quantidadeInstalada/meta)*100;
When you're dividing int by int (as in "quantidadeInstalada/meta"), you'll get an int. Either use a fractional type (e.g. decimal, double) from the very beginning, as Shaun suggested, or (if the integral types have to stay), cast the values to a fractional type in the division expression (as itsme86 shown).
This question already has answers here:
How do I round to the nearest 0.5?
(10 answers)
Closed 9 years ago.
I'm trying to round some values as the following examples and need some help to write the math calculation for it:
input -> 25 ÷ 4 = 6.25 output -> 6.5
input -> 15.5 ÷ 4 = 3.875 output -> 4.0
input -> 24.5 ÷ 4 = 6.125 output -> 6.0
any idea how to write the round math procedure please?!
This should do it.
double Divide(double numerator, double denominator)
{
double result = numerator / denominator;
//round to nearest half-integer
result = Math.Round(result * 2, MidpointRounding.AwayFromZero) / 2;
// due to peculiarities of IEEE754 floating point arithmetic
// we need to round again after dividing back by two
// to avoid a result like 1.49999999.
return Math.Round(result, 1);
}
Sorry for not aware the difficulty you encountered, so I guess maybe the different types of floating point number. Following code just does that:
public static decimal RoundedDivide<T>(T a, T b) {
var x=2*Convert.ToDecimal(a)/Convert.ToDecimal(b);
x=((int)(.5m+x)>x?1:0)+(int)x;
return x/2;
}
Two things to note:
I cannot bound a constraint of ValueType, so if you pass objects of reference types, it may throw
If you wish double as the return type, then just change it and also change .5m to .5, Convert.ToDecimal to Convert.ToDouble
I have the following code :
double a = 8/ 3;
Response.Write(a);
It returns the value 2. Why? I need at least one decimal digit. Something like 2.6, or 2.66. How can I get such results?
Try
double a = 8/3.0d;
or
double a = 8.0d/3;
to get a precise answer.
Since in expression a = 8/3 both the operands are int so the result is int irrespective of the fact that it is being stored in a double. The results are always in the higher data type of operands
EDIT
To answer
8 and 3 are get from variable. Can I do a sort of cast?
In case the values are coming from a variable you can cast one of the operands into double like:
int b = 8;
int c = 3;
double a = ((double) b) /c;
Because the calculation are being done in integer type not double. To make it double use:
double a = 8d/ 3d;
Response.Write(a);
Or
double a = 8.0/ 3.0;
Response.Write(a);
One of your operands should be explicitly marked as double either by using d or specifying a decimal point 0
or if you need you can cast them to double before the calculations. You can cast either one or both operands to double.
double a = ((double) 8)/((double)3)
because 8 and 3 are integer numbers and interpreter rounds it to 2.
You can simply advise to interpreter that you numbers are floating numbers:
double a = (double)8 / 3;
Because its making a rounding towards minus, its the way its implemented in the framework. However if you specify the precision by using the above example:
double a = 8/3.0d;
then rounding is no longer performed.
Or in simple terms you assigned an integer value to a double, thats why the rounding was performed in the first place. It saw an operation with integers.
Coz 8 and 3 both ints. And int's division operator with two ints in it returns int as well. (F12 when the cursor is on slash sign).