This question already has answers here:
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 3 years ago.
I am grabbing total RAM of a computer system and available RAM and trying to work out what percentage is available.
I am using the following code:
double percent = my.Info.AvailablePhysicalMemory / my.Info.TotalPhysicalMemory;
and have also tried:
decimal percent = my.Info.AvailablePhysicalMemory / my.Info.TotalPhysicalMemory;
I am sure it's an issue with the type but I am unsure why both methods give a result of 0.
The actual values are Total: 17072574464 and Available: 8746000384. The values come back from the system cast as ulong. So what does percent always equal 0? If I put the numbers in directly it works fine. Just can't use the ulong variables - hence why I am sure it's my lack of experience with types in C# that is the problem.
You are trying to divide an integer by an integer, which always rounds down. You need to convert to a floating point number before you divide; for example:
double percent = my.Info.AvailablePhysicalMemory * 1.0 / my.Info.TotalPhysicalMemory;
Related
This question already has answers here:
Why does integer division in C# return an integer and not a float?
(8 answers)
Closed 6 years ago.
I am really confused why this is happening, my code:
double x = Math.Sqrt(2/3);
MessageBox.Show(x.ToString());
Displays 0.
The answer is
0.8164, I know I will also have to use Math.Round to round this up, but for the moment the issue is I'm getting 0
The problem is caused by automatic integer evaluation of the numbers. Use:
double x = Math.Sqrt(2f/3f);
MessageBox.Show(x.ToString());
2 / 3 is an integer operation, what you want is 2.0 / 3 which means I want to use floating point numbers.
What you consider an Intereger is different from what you know from Maths. In programming languages it means that a result of an int-operation is allways an integer in itself.
In your example 2 / 3 is an integer-operation which means the result is rounded down to the nearest integer, which is zero. To avoid this indicate that at least one of your operands should be treates as some floating-point value, either using 2.0 or 2f (alternativly 3.0 or 3f).
This question already has answers here:
C# is rounding down divisions by itself
(10 answers)
Closed 8 years ago.
I am working in Unity3D with C# and I get a weird result. Can anyone tell me why my code equals 0?
float A = 1 / 90;
The literals 1 and 90 are interpreted as an int. So integer division is used. After that the result is converted to a float.
In general C# will read all sequences (without decimal dot) of digits as an int. An int will be converted to a float if necessary. But before the assignment, that's not necessary. So all calculations in between are done as ints.
In other words, what you've written is:
float A = (float) ((int) 1)/((int) 90)
(made it explicit here, this is more or less what the compiler reads).
Now a division of two int's is processed such that it takes only the integral part into account. The integral part of 0.011111 is 0 thus zero.
If you however modify one of the literals to a floating point (1f, 1.0f, 90f,...) or both, this will work. Thus use one of these:
float A = 1/90.0f;
float A = 1.0f/90;
float A = 1.0f/90.0f;
In that case, floating point division will be performed. Which takes into account both parts.
etc.
This question already has answers here:
Why returns C# Convert.ToDouble(5/100) 0.0 and not 0.05
(7 answers)
Closed 8 years ago.
I have a double that when assigned is always coming back as 0.0, but should not happen.
The code I have is
double test = (27096140 / 27216140);
The answer to this should be 0.9955....
I am not using the double primitive type correctly?
Cheers
Edit
Of course, I was using int and no floating points.
Knew it would be a matter of common sense.
You are using integer division, you should use floating-point division:
double test = (27096140.0 / 27216140);
You are actually performing integer division here because the 2 numbers are whole. That forces the result of the calculation to be int which truncates the precision so you end up with 0.
You need to tell the compiler you are actually working with floating-point numbers and not int's, just adding a floating point to one of the numbers should fix this e.g.
double test = (27096140.0 / 27216140);
Alternatively, you could actually declare the numbers as doubles
double a = 27096140;
double b = 27216140;
double test = (a / b);
Or even cast the number in the calculation
double test = ((double)27096140 / 27216140)
Beacuse you're using integers instead of doubles. Use:
double test = (27096140.0d / 27216140.0d);
The numbers 27096140 and 27216140 are ints, so the result of the division is also an int: 0.
Then this is cast to a double: 0.0
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
What is the difference between Decimal, Float and Double in C#?
Today I'm wondering about Double in .Net. I've used it with Int32 the past days and started wondering what the max value is.
The MSDN page for Double.MaxValue says 1.7976931348623157E+308. I'm pretty sure I'm reading that wrong.
How many bytes does Double take up (in memory)?
What is the actual maximum number (explain the the E+308)?
Is Double.MaxValue bigger than UInt32? Bigger than UInt64?
And while we are at it, what is the difference between Float and Double?
Basically,
Double is 64 bit floating point value and float is a 32 bit.
So double is able to store twice big value as of float.
http://msdn.microsoft.com/en-us/library/678hzkk9(v=vs.80).aspx
http://msdn.microsoft.com/en-us/library/b1e65aza(v=vs.71).aspx
Just read the top lines on the links, you'll get an idea.
About E+308: though 2^64 is far less that 1e+308, you must consider that double is not "precise" number, it has only a few significant digits (precision), so it does not need to store all ~308 digits. With this logic behind the double structure, it can contain numbers up to e+308 in 64 bits.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
c# - How do I round a decimal value to 2 decimal places (for output on a page)
i have value of 1.564 and 1.565 i need round this value to 1.56 and 1.56,which function suitable for this in c#.
do a multiply by 100 followed by a floor and followed by a divide by hundred. I am sure that there is a better way of doing it though
Math.floor(n*100)/100
To remove the less significant digits (1.348 -> 1.34):
Math.Floor(number * 100) / 100;
To round the number to two decimals:
Math.Round(number, 2);
To represent it as a string, for display:
number.ToString("#.00");