How come dividing two 32 bit int numbers as ( int / int ) returns to me 0, but if I use Decimal.Divide() I get the correct answer? I'm by no means a c# guy.
int is an integer type; dividing two ints performs an integer division, i.e. the fractional part is truncated since it can't be stored in the result type (also int!). Decimal, by contrast, has got a fractional part. By invoking Decimal.Divide, your int arguments get implicitly converted to Decimals.
You can enforce non-integer division on int arguments by explicitly casting at least one of the arguments to a floating-point type, e.g.:
int a = 42;
int b = 23;
double result = (double)a / b;
In the first case, you're doing integer division, so the result is truncated (the decimal part is chopped off) and an integer is returned.
In the second case, the ints are converted to decimals first, and the result is a decimal. Hence they are not truncated and you get the correct result.
The following line:
int a = 1, b = 2;
object result = a / b;
...will be performed using integer arithmetic. Decimal.Divide on the other hand takes two parameters of the type Decimal, so the division will be performed on decimal values rather than integer values. That is equivalent of this:
int a = 1, b = 2;
object result = (Decimal)a / (Decimal)b;
To examine this, you can add the following code lines after each of the above examples:
Console.WriteLine(result.ToString());
Console.WriteLine(result.GetType().ToString());
The output in the first case will be
0
System.Int32
..and in the second case:
0,5
System.Decimal
I reckon Decimal.Divide(decimal, decimal) implicitly converts its 2 int arguments to decimals before returning a decimal value (precise) where as 4/5 is treated as integer division and returns 0
You want to cast the numbers:
double c = (double)a/(double)b;
Note: If any of the arguments in C# is a double, a double divide is used which results in a double. So, the following would work too:
double c = (double)a/b;
here is a Small Program :
static void Main(string[] args)
{
int a=0, b = 0, c = 0;
int n = Convert.ToInt16(Console.ReadLine());
string[] arr_temp = Console.ReadLine().Split(' ');
int[] arr = Array.ConvertAll(arr_temp, Int32.Parse);
foreach (int i in arr)
{
if (i > 0) a++;
else if (i < 0) b++;
else c++;
}
Console.WriteLine("{0}", (double)a / n);
Console.WriteLine("{0}", (double)b / n);
Console.WriteLine("{0}", (double)c / n);
Console.ReadKey();
}
In my case nothing worked above.
what I want to do is divide 278 by 575 and multiply by 100 to find percentage.
double p = (double)((PeopleCount * 1.0 / AllPeopleCount * 1.0) * 100.0);
%: 48,3478260869565 --> 278 / 575 ---> 0
%: 51,6521739130435 --> 297 / 575 ---> 0
if I multiply the PeopleCount by 1.0 it makes it decimal and division will be 48.34...
also multiply by 100.0 not 100.
If you are looking for 0 < a < 1 answer, int / int will not suffice. int / int does integer division. Try casting one of the int's to a double inside the operation.
The answer marked as such is very nearly there, but I think it is worth adding that there is a difference between using double and decimal.
I would not do a better job explaining the concepts than Wikipedia, so I will just provide the pointers:
floating-point arithmetic
decimal data type
In financial systems, it is often a requirement that we can guarantee a certain number of (base-10) decimal places accuracy. This is generally impossible if the input/source data is in base-10 but we perform the arithmetic in base-2 (because the number of decimal places required for the decimal expansion of a number depends on the base; one third takes infinitely many decimal places to express in base-10 as 0.333333..., but it takes only one decimal in base-3: 0.1).
Floating-point numbers are faster to work with (in terms of CPU time; programming-wise they are equally simple) and preferred whenever you want to minimize rounding error (as in scientific applications).
Related
Does anyone know why integer division in C# returns an integer and not a float?
What is the idea behind it? (Is it only a legacy of C/C++?)
In C#:
float x = 13 / 4;
//== operator is overridden here to use epsilon compare
if (x == 3.0)
print 'Hello world';
Result of this code would be:
'Hello world'
Strictly speaking, there is no such thing as integer division (division by definition is an operation which produces a rational number, integers are a very small subset of which.)
While it is common for new programmer to make this mistake of performing integer division when they actually meant to use floating point division, in actual practice integer division is a very common operation. If you are assuming that people rarely use it, and that every time you do division you'll always need to remember to cast to floating points, you are mistaken.
First off, integer division is quite a bit faster, so if you only need a whole number result, one would want to use the more efficient algorithm.
Secondly, there are a number of algorithms that use integer division, and if the result of division was always a floating point number you would be forced to round the result every time. One example off of the top of my head is changing the base of a number. Calculating each digit involves the integer division of a number along with the remainder, rather than the floating point division of the number.
Because of these (and other related) reasons, integer division results in an integer. If you want to get the floating point division of two integers you'll just need to remember to cast one to a double/float/decimal.
See C# specification. There are three types of division operators
Integer division
Floating-point division
Decimal division
In your case we have Integer division, with following rules applied:
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. The result is zero
or positive when the two operands have the same sign and zero or
negative when the two operands have opposite signs.
I think the reason why C# use this type of division for integers (some languages return floating result) is hardware - integers division is faster and simpler.
Each data type is capable of overloading each operator. If both the numerator and the denominator are integers, the integer type will perform the division operation and it will return an integer type. If you want floating point division, you must cast one or more of the number to floating point types before dividing them. For instance:
int x = 13;
int y = 4;
float x = (float)y / (float)z;
or, if you are using literals:
float x = 13f / 4f;
Keep in mind, floating points are not precise. If you care about precision, use something like the decimal type, instead.
Since you don't use any suffix, the literals 13 and 4 are interpreted as integer:
Manual:
If the literal has no suffix, it has the first of these types in which its value can be represented: int, uint, long, ulong.
Thus, since you declare 13 as integer, integer division will be performed:
Manual:
For an operation of the form x / y, binary operator overload resolution is applied to select a specific operator implementation. The operands are converted to the parameter types of the selected operator, and the type of the result is the return type of the operator.
The predefined division operators are listed below. The operators all compute the quotient of x and y.
Integer division:
int operator /(int x, int y);
uint operator /(uint x, uint y);
long operator /(long x, long y);
ulong operator /(ulong x, ulong y);
And so rounding down occurs:
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. The result is zero or positive when the two operands have the same sign and zero or negative when the two operands have opposite signs.
If you do the following:
int x = 13f / 4f;
You'll receive a compiler error, since a floating-point division (the / operator of 13f) results in a float, which cannot be cast to int implicitly.
If you want the division to be a floating-point division, you'll have to make the result a float:
float x = 13 / 4;
Notice that you'll still divide integers, which will implicitly be cast to float: the result will be 3.0. To explicitly declare the operands as float, using the f suffix (13f, 4f).
Might be useful:
double a = 5.0/2.0;
Console.WriteLine (a); // 2.5
double b = 5/2;
Console.WriteLine (b); // 2
int c = 5/2;
Console.WriteLine (c); // 2
double d = 5f/2f;
Console.WriteLine (d); // 2.5
It's just a basic operation.
Remember when you learned to divide. In the beginning we solved 9/6 = 1 with remainder 3.
9 / 6 == 1 //true
9 % 6 == 3 // true
The /-operator in combination with the %-operator are used to retrieve those values.
The result will always be of type that has the greater range of the numerator and the denominator. The exceptions are byte and short, which produce int (Int32).
var a = (byte)5 / (byte)2; // 2 (Int32)
var b = (short)5 / (byte)2; // 2 (Int32)
var c = 5 / 2; // 2 (Int32)
var d = 5 / 2U; // 2 (UInt32)
var e = 5L / 2U; // 2 (Int64)
var f = 5L / 2UL; // 2 (UInt64)
var g = 5F / 2UL; // 2.5 (Single/float)
var h = 5F / 2D; // 2.5 (Double)
var i = 5.0 / 2F; // 2.5 (Double)
var j = 5M / 2; // 2.5 (Decimal)
var k = 5M / 2F; // Not allowed
There is no implicit conversion between floating-point types and the decimal type, so division between them is not allowed. You have to explicitly cast and decide which one you want (Decimal has more precision and a smaller range compared to floating-point types).
As a little trick to know what you are obtaining you can use var, so the compiler will tell you the type to expect:
int a = 1;
int b = 2;
var result = a/b;
your compiler will tell you that result would be of type int here.
In my C# application I want to implement a simple calculation. I got this code:
private void button1_Click(object sender, EventArgs e)
{
int percentField;
int priceField;
int result;
percentField = int.Parse(txtPercentNew.Text);
priceField = int.Parse(txtPriceNew.Text);
result = priceField / 100 * percentField;
MessageBox.Show(result.ToString());
}
But the problem is the MessageBox displays me 0. I can't figure out why.
Can someone please give me a hint what I am doing wrong?
Your variables are integers, which means that / performs integer division. Unless priceField is at least equal to 100 you will always get 0 as the result.
You can correct the problem by casting priceField to a floating point type before dividing:
(double)priceField / 100 * percentField;
However, this will not work while result is of type int because the compiler wants to protect you from inadvertent rounding errors. So you either have to cast back to an integer (losing precision due to rounding):
result = (int)((double)priceField / 100 * percentField);
or else make result be a double as well.
You are using integers instead of floating point numbers.
As a consequence, rounding off occurs during calculation.
Use float or double instead of int.
Probably your priceField is less then 100 and since you doing integer division, it creates 0 as a result.
From / Operator (C# Reference)
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, as the following example shows.
Just cast one of your variables to floating point type like;
result = priceField / 100d * percentField;
or
result = (double)priceField / 100 * percentField;
You are working with only integers, try
double result;
result = priceField / (double)100 * percentField;
In your code, you are dividing by 100. Which means every int you are going to divide less than 100 will result in a value between [0 - 1]. When implicitly casting to an int, the result will be floored. Therefor, a 0.1 will become 0 - a 0.9 will become 0 - ...
try cast to double, because you're working with integers it results in 0.
example here
The problem is you're using int for each value.
Change result to a double and try this:
result = (double)priceField / 100 * percentField;
It should work; however, if you want to do this properly I recommend you read about MidpointRounding.
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).
Does anyone know why integer division in C# returns an integer and not a float?
What is the idea behind it? (Is it only a legacy of C/C++?)
In C#:
float x = 13 / 4;
//== operator is overridden here to use epsilon compare
if (x == 3.0)
print 'Hello world';
Result of this code would be:
'Hello world'
Strictly speaking, there is no such thing as integer division (division by definition is an operation which produces a rational number, integers are a very small subset of which.)
While it is common for new programmer to make this mistake of performing integer division when they actually meant to use floating point division, in actual practice integer division is a very common operation. If you are assuming that people rarely use it, and that every time you do division you'll always need to remember to cast to floating points, you are mistaken.
First off, integer division is quite a bit faster, so if you only need a whole number result, one would want to use the more efficient algorithm.
Secondly, there are a number of algorithms that use integer division, and if the result of division was always a floating point number you would be forced to round the result every time. One example off of the top of my head is changing the base of a number. Calculating each digit involves the integer division of a number along with the remainder, rather than the floating point division of the number.
Because of these (and other related) reasons, integer division results in an integer. If you want to get the floating point division of two integers you'll just need to remember to cast one to a double/float/decimal.
See C# specification. There are three types of division operators
Integer division
Floating-point division
Decimal division
In your case we have Integer division, with following rules applied:
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. The result is zero
or positive when the two operands have the same sign and zero or
negative when the two operands have opposite signs.
I think the reason why C# use this type of division for integers (some languages return floating result) is hardware - integers division is faster and simpler.
Each data type is capable of overloading each operator. If both the numerator and the denominator are integers, the integer type will perform the division operation and it will return an integer type. If you want floating point division, you must cast one or more of the number to floating point types before dividing them. For instance:
int x = 13;
int y = 4;
float x = (float)y / (float)z;
or, if you are using literals:
float x = 13f / 4f;
Keep in mind, floating points are not precise. If you care about precision, use something like the decimal type, instead.
Since you don't use any suffix, the literals 13 and 4 are interpreted as integer:
Manual:
If the literal has no suffix, it has the first of these types in which its value can be represented: int, uint, long, ulong.
Thus, since you declare 13 as integer, integer division will be performed:
Manual:
For an operation of the form x / y, binary operator overload resolution is applied to select a specific operator implementation. The operands are converted to the parameter types of the selected operator, and the type of the result is the return type of the operator.
The predefined division operators are listed below. The operators all compute the quotient of x and y.
Integer division:
int operator /(int x, int y);
uint operator /(uint x, uint y);
long operator /(long x, long y);
ulong operator /(ulong x, ulong y);
And so rounding down occurs:
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. The result is zero or positive when the two operands have the same sign and zero or negative when the two operands have opposite signs.
If you do the following:
int x = 13f / 4f;
You'll receive a compiler error, since a floating-point division (the / operator of 13f) results in a float, which cannot be cast to int implicitly.
If you want the division to be a floating-point division, you'll have to make the result a float:
float x = 13 / 4;
Notice that you'll still divide integers, which will implicitly be cast to float: the result will be 3.0. To explicitly declare the operands as float, using the f suffix (13f, 4f).
Might be useful:
double a = 5.0/2.0;
Console.WriteLine (a); // 2.5
double b = 5/2;
Console.WriteLine (b); // 2
int c = 5/2;
Console.WriteLine (c); // 2
double d = 5f/2f;
Console.WriteLine (d); // 2.5
It's just a basic operation.
Remember when you learned to divide. In the beginning we solved 9/6 = 1 with remainder 3.
9 / 6 == 1 //true
9 % 6 == 3 // true
The /-operator in combination with the %-operator are used to retrieve those values.
The result will always be of type that has the greater range of the numerator and the denominator. The exceptions are byte and short, which produce int (Int32).
var a = (byte)5 / (byte)2; // 2 (Int32)
var b = (short)5 / (byte)2; // 2 (Int32)
var c = 5 / 2; // 2 (Int32)
var d = 5 / 2U; // 2 (UInt32)
var e = 5L / 2U; // 2 (Int64)
var f = 5L / 2UL; // 2 (UInt64)
var g = 5F / 2UL; // 2.5 (Single/float)
var h = 5F / 2D; // 2.5 (Double)
var i = 5.0 / 2F; // 2.5 (Double)
var j = 5M / 2; // 2.5 (Decimal)
var k = 5M / 2F; // Not allowed
There is no implicit conversion between floating-point types and the decimal type, so division between them is not allowed. You have to explicitly cast and decide which one you want (Decimal has more precision and a smaller range compared to floating-point types).
As a little trick to know what you are obtaining you can use var, so the compiler will tell you the type to expect:
int a = 1;
int b = 2;
var result = a/b;
your compiler will tell you that result would be of type int here.
Im trying to write something to get my images to show correctly.
I have 2 numbers "breedtePlaatje" and "hoogtePlaatje". When i load those 2 vars with the values i get back "800" and "500" i expect "verH" to be (500 / 800) = 0,625. Tho the value of verH = 0..
This is the code:
int breedtePlaatje = Convert.ToInt32(imagefield.Width);
int hoogtePlaatje = Convert.ToInt32(imagefield.Height);
//Uitgaan van breedte plaatje
if (breedtePlaatje > hoogtePlaatje)
{
double verH = (hoogtePlaatje/breedtePlaatje);
int vHeight = Convert.ToInt32(verH * 239);
mOptsMedium.Height = vHeight;
mOptsMedium.Width = 239;
//Hij wordt te klein en je krijgt randen te zien, dus plaatje zelf instellen
if (hoogtePlaatje < 179)
{
mOptsMedium.Height = 179;
mOptsMedium.Width = 239;
}
}
Any tips regarding my approach would be lovely aswell.
Dividing int by int gives an int.
double verH = (hoogtePlaatje/breedtePlaatje);
The right hand side of the assignment is an integer value.
Change breedtePlaatje and/or hoogtePlaatje to double and you will get the answer you expect.
Integer division will result in an Integer being returned as the division result.
You need one of the parameters of the division to be a float in order for the result to be a float. You can do this by casting one of them to a float.
double verH = (double)hoogtePlaatje/breedtePlaatje;
Or
double verH = hoogtePlaatje/(double)breedtePlaatje;
See the C# spec regarding division.
When you divide two integers, C# uses integer division, where the fractional part is discarded. In your case you're getting:
500 / 800 = 0 + 5/8
Which, discarding the fractional part, gives:
500 / 800 = 0
To get floating point division, cast one of the arguments to either double, float or decimal depending on the level of precision you need, which will cause the other argument to be implicitly converted to the same type and the division carried out using floating point rules instead of integer rules, e.g.
double result = (double)breedtePlaatje / hoogtePlaatje ;
I have never used C#, but probably you will need to cast one of the variables to double, like this:
double verH = (double)hoogtePlaatje/breedtePlaatje;
Try this:
double verH = double (hoogtePlaatje) / breedtePlaateje;
If you divide an int by an int, you will get a truncated answer. Cast one of them up to a double, and the entire division will be done as double.