
c# - What the difference between nint vs int? - Stack Overflow
2021年11月10日 · I'm curious about the introduction of nint for native-sized integers in C# 9. Could someone provide a detailed comparison between int and nint, focusing on their practical differences in usage and behavior?
.net - What are Native-sized integers in C# 9.0? - Stack Overflow
2021年6月11日 · The first three lines of @Sommmen's link says it all: "Starting in C# 9.0, you can use the nint and nuint keywords to define native-sized integers. These are 32-bit integers when running in a 32-bit process, or 64-bit integers when running in a 64-bit process. They can be used for interop scenarios, low-level libraries, and to optimize …
Fortran - want to round to one decimal point - Stack Overflow
2016年8月23日 · The `round_x = nint (x*10d0)/10d0' operator rounds x (for abs (x) < 2**31/10, for large numbers use dnint ()) and assigns the rounded value to the round_x variable for further calculations. As mentioned in the answers above, not all numbers with one significant digit after the decimal point have an exact representation, for example, 0.3 does not.
System.IntPtr and nint in C# Microsoft.CodeAnalysis
2021年5月20日 · nint is a C# v9 keyword that maps to IntPtr, the type that the runtime understands. Just like int maps to Int32 and string to String, etc. It is indeed not a simple alias like int, they removed the dangerous members like ToInt32 (), ToPointer (), etc. Having code analysis represent it in SpecialType is sensible, except that it takes an act of God to modify the core types. You can ask Him at ...
How does Fortran convert a real number to Integer
2017年8月25日 · Either use NINT() which is the nearest integer, or INT(). INT() only returns the signed integer part of a number. NINT() works as follows: If a is greater than zero, NINT (a) has the value INT (a+ 0.5); if a is less than or equal to zero, NINT (a) has the value INT (a- 0.5). Specifically NINT(0.5d0) = 1
Fortran 77 Real to Int rounding Direction? - Stack Overflow
Only as completion to the existing answers I want to add an example how the commercial rounds can be realized without using NINT by L = INT(F + 0.5) where L is INTEGER and F is a positive REAL number. I've found this in FORTRAN 77 code samples from the last century. Extending this to negative REAL numbers by L = SIGN(1.0,F)*INT(ABS(F) + 0.5)
c# - How to declare a IntPtr? - Stack Overflow
2010年3月15日 · In addition to constant-like static-readonly IntPtr/UIntPtr variables, you can use nint and nuint which are backed by IntPtr and UIntPtr, respectively. These types can have integer and unsigned integer values assigned to them and can also be defined as constants.
How can I pass a pointer to an integer in C# - Stack Overflow
2014年2月12日 · One option is simply to use C# pointer types - this requires unsafe block (or modifier on method/class), and compiling with /unsafe: [DllImport(...)] static extern int GetBuffer(byte* buffer, ref int maxSize); Buffer can be allocated in several different ways. One would be to use a pinned heap array: fixed (byte* buffer = new byte[4096]) { int maxSize = buffer.Length; GetBuffer(buffer, ref ...
unsafe - Is it possible to reinterpret an object reference as native ...
2022年6月24日 · Disclaimer: I understand this is an unsafe operation and bad practice, I merely want to know if it's possible. Basically, I'm trying to convert an object reference to a nint (IntPtr). They take up ...
CEILING and FLOOR function in Fortran - Stack Overflow
2015年3月31日 · Those variables are real because of implicit typing. Contrast this with nintx which is indeed an integer variable. List-directed output (the write(*,*) part) has as natural result formatting the real variables as you see. If you instead directly print the function results write(*,*) NINT(x), CEILING(X), FLOOR(X) you will be less surprised.