what is equal to the c++ size_t in c# - c#

I have a struct in c++:
struct some_struct{
uchar* data;
size_t size;
}
I want to pass it between manged(c#) and native(c++). What is the equivalent of size_t in C# ?
P.S. I need an exact match in the size because any byte difference will results in huge problem while wrapping
EDIT:
Both native and manged code are under my full control ( I can edit whatever I want)

There is no C# equivalent to size_t.
The C# sizeof() operator always returns an int value regardless of platform, so technically the C# equivalent of size_t is int, but that's no help to you.
(Note that Marshal.SizeOf() also returns an int.)
Also note that no C# object can be larger than 2GB in size as far as sizeof() and Marshal.Sizeof() is concerned. Arrays can be larger than 2GB, but you cannot use sizeof() or Marshal.SizeOf() with arrays.
For your purposes, you will need to know what the version of code in the DLL uses for size_t and use the appropriate size integral type in C#.
One important thing to realise is that in C/C++ size_t will generally have the same number of bits as intptr_t but this is NOT guaranteed, especially for segmented architectures.
I know lots of people say "use UIntPtr", and that will normally work, but it's not GUARANTEED to be correct.
From the C/C++ definition of size_t, size_t
is the unsigned integer type of the result of the sizeof operator;

The best equivalent for size_t in C# is the UIntPtr type. It's 32-bit on 32-bit platforms, 64-bit on 64-bit platforms, and unsigned.

You better to use nint/nuint which is wrapper around IntPtr/UIntPtr

Related

About the "GetBytes" implementation in BitConverter

I've found that the implementation of the GetBytes function in .net framework is something like:
public unsafe static byte[] GetBytes(int value)
{
byte[] bytes = new byte[4];
fixed(byte* b = bytes)
*((int*)b) = value;
return bytes;
}
I'm not so sure I understand the full details of these two lines:
fixed(byte* b = bytes)
*((int*)b) = value;
Could someone provide a more detailed explanation here? And how should I implement this function in standard C++?
Could someone provide a more detailed explanation here?
The MSDN documentation for fixed comes with numerous examples and explanation -- if that's not sufficient, then you'll need to clarify which specific part you don't understand.
And how should I implement this function in standard C++?
#include <cstring>
#include <vector>
std::vector<unsigned char> GetBytes(int value)
{
std::vector<unsigned char> bytes(sizeof(int));
std::memcpy(&bytes[0], &value, sizeof(int));
return bytes;
}
Fixed tells the garbage collector not to move a managed type so that you can access that type with standard pointers.
In C++, if you're not using C++/CLI (i.e. not using .NET) then you can just use a byte-sized pointer (char) and loop through the bytes in whatever you're trying to convert.
Just be aware of endianness...
First fixed has to be used because we want to assign a pointer to a managed variable:
The fixed statement prevents the garbage collector from relocating a
movable variable. The fixed statement is only permitted in an unsafe
context. Fixed can also be used to create fixed size buffers.
The fixed statement sets a pointer to a managed variable and "pins"
that variable during the execution of the statement. Without fixed,
pointers to movable managed variables would be of little use since
garbage collection could relocate the variables unpredictably. The
C# compiler only lets you assign a pointer to a managed variable in a
fixed statement. Ref.
Then we declare a pointer to byte and assign to the start of the byte array.
Then, we cast the pointer to byte to a pointer to int, dereference it and assign it to the int passed in.
The function creates a byte array that contains the same binary data as your platform's representation of the integer value. In C++, this can be achieved (for any type really) like so:
int value; // or any type!
unsigned char b[sizeof(int)];
unsigned char const * const p = reinterpret_cast<unsigned char const *>(&value);
std::copy(p, p + sizeof(int), b);
Now b is an array of as many bytes as the size of the type int (or whichever type you used).
In C# you need to say fixed to obtain a raw pointer, since usually you do not have raw pointers in C# on account of objects not having a fixed location in memory -- the garbage collector can move them around at any time. fixed prevents this and fixes the object in place so a raw pointer can make sense.
You can implement GetBytes() for any POD type with a simple function template.
#include <vector>
template <typename T>
std::vector<unsigned char> GetBytes(T value)
{
return std::vector<unsigned char>(reinterpret_cast<unsigned char*>(&value),
reinterpret_cast<unsigned char*>(&value) + sizeof(value));
}
Here is a C++ header-only library that may be of help.
BitConverter
The idea of implementing the GetBytes function in C++ is straight-forward: compute each byte of the value according to specified layout. For example, let's say we need to get the bytes of an unsigned 16-bit integer in big endian. We can divide the value by 256 to get the first byte, and take the remainder as the second byte.
For floating-point numbers, the algorithm is a little bit more complicated. We need to get the sign, exponent, and mantissa of the number, and encode them as bytes. See https://en.wikipedia.org/wiki/Double-precision_floating-point_format

Can someone tell me what this crazy c++ statement means in C#?

First off, no I am not a student...just a C# guy porting a C++ library.
What do these two crazy lines mean? What are they equivalent to in C#? I'm mostly concerned with the size_t and sizeof. Not concerned about static_cast or assert..I know how to deal with those.
size_t Index = static_cast<size_t>((y - 1620) / 2);
assert(Index < sizeof(DeltaTTable)/sizeof(double));
y is a double and DeltaTTable is a double[]. Thanks in advance!
size_t is a typedef for an unsigned integer type. It is used for sizes of things, and may be 32 or 64 bits in size. The particular size of a size_t is implementation defined, but it is unsigned.
I suppose in C# you could use a 64-bit unsigned integer type.
All sizeof does is return the size in bytes of a C++ type. Every type takes up a certain quantity of room, and sizeof returns that size.
What your code is doing is computing the number of doubles (64-bit floats) that the DeltaTTable takes up. Essentially, it's ensuring that the table is larger than some size based on y, whatever that is.
There is no equivalent of sizeof in C#, nor does it need it. There is no reason for you to port this code to C#.
The bad news first you can't do that in C#. There's no static cast only dynamic casts. However the good news is it doesn't matter.
The two lines of code is asserting that the index is in bounds of the table so that the code won't accidentally read some arbitrary memory location. The CLR takes care of that for you. So when porting just ignore those lines they are automatically there for you any ways.
Of course this is based on an assumption based on the pattern of the code. There's no information on what Y represents and how Index is used
sizeOf calculates how much memory in bytes the DeltaTable type takes.
There is not equivalent to calculate the size like this in c# AFAIK.
I guess size_t much be a struct type in C++ code.

is java byte the same as C# byte?

Native method from dll works in java if the input parameter is array of bytes - byte[].
If we use the same method from c# it throws EntryPointNotFoundException.
Is that because of byte[] in java and c# are different things? and if it's so how should I use native function from c#?
Java lacks the unsigned types. In particular, Java lacks a primitive type for an unsigned byte. The Java byte type is signed, while the C# byte is unsigned and sbyte is signed.
Is that because of byte[] in java and c# are different things?
Yes.
Endianness: Java stores things internally as Big Endian, while .NET is Little Endian by default.
Signedness: C# bytes are unsigned. Java bytes are signed.
See different results when converting int to byte array - .NET vs Java.
What's the signature of the native function? How do you declare it in Java and in C#?
The most common reason for EntryPointNotFoundException is that function name is mangled (esp. true if function is written in C++) or misspelled.
Another source of problem is 'W' and 'A' suffixes for WinAPI function used to distinguish ANSI and Unicode versions of functions. .NET interop mechanism can try to guess the function suffix, so that may be the source of confusion,
Java Byte:
java byte: The byte data type is an 8-bit signed two's complement integer. It has a minimum value of -128 and a maximum value of 127 (inclusive). The byte data type can be useful for saving memory in large arrays, where the memory savings actually matters. They can also be used in place of int where their limits help to clarify your code; the fact that a variable's range is limited can serve as a form of documentation.
more for Java Byte
C# Byte
Byte Represents an 8-bit unsigned integer,Byte is an immutable value type that represents unsigned integers with values that range from 0 (which is represented by the Byte.MinValue constant) to 255 (which is represented by the Byte.MaxValue constant). The .NET Framework also includes a signed 8-bit integer value type, SByte, which represents values that range from -128 to 127.
more for c# Byte

Why are the unsigned CLR types so difficult to use in C#?

I came from a mostly C/C++ background before I began using C#. One of the things I did with my first project in C# was make a class like this
class Element{
public uint Size;
public ulong BigThing;
}
I was then mortified by the fact that this requires casting:
int x=MyElement.Size;
as does
int x=5;
uint total=MyElement.Size+x;
Why did the language designers decide to make the signed and unsigned integer types not implicitly castable? And why are the unsigned types not used more throughout the .Net library? For instance String.Length can never be negative, yet it is a signed integer.
Why did the language designers decide to make the signed and unsigned integer types not
implicitly castable?
Because that could lose data or throw any exception, neither of which is generally a good thing to allow implicitly. (The implicit conversion from long to double can lose data too, admittedly, but in a different way.)
And why are the unsigned types not used more throughout the .Net library
Unsigned types aren't CLS-compliant - not all .NET languages have always supported them. For example, Visual Basic didn't have "native" support for unsigned data types in .NET 1.0 and 1.1; it was added to the language for 2.0. (You could still use them, but they weren't part of the language itself - you couldn't use the normal arithmetic operators, for example.)
Along with Jon's answer, just because an unsigned number can't be negative doesn't mean it isn't bigger than a signed one. uint is 0 to 4,294,967,295 but int is -2,147,483,648 to 2,147,483,647. Plenty of room above int's max for loss.
Because implicitly converting an unsigned integer of 3B into an signed integer is going to blow up.
Unsigned has twice the maximum value of signed. It's the same reason you can't cast a long to an int.
I was then mortified by the fact that this requires casting:
int x=MyElement.Size;
But you are contradicting yourself here. If you really (really) need Size to be unsigned than assigning it to (signed) x is an error. A deep flaw in your code.
For instance String.Length can never be negative, yet it is a signed integer
But String.IndexOf can return a negative number, and it would be awkward if String.Length and Index values where of different types.
And while in theory there would be merit in an unsigned String.Length (4 GB cap), in practice even the current 2GB is large enough (because strings of that length are rare and unworkable anyway).
So the real answer is: Why use unsigned in the first place?
On the second count: because they wanted the CLR to be compatible with languages that don't have unsigned datatypes (read: VB.NET).

Is an int a 64-bit integer in 64-bit C#?

In my C# source code I may have declared integers as:
int i = 5;
or
Int32 i = 5;
In the currently prevalent 32-bit world they are equivalent. However, as we move into a 64-bit world, am I correct in saying that the following will become the same?
int i = 5;
Int64 i = 5;
No. The C# specification rigidly defines that int is an alias for System.Int32 with exactly 32 bits. Changing this would be a major breaking change.
The int keyword in C# is defined as an alias for the System.Int32 type and this is (judging by the name) meant to be a 32-bit integer. To the specification:
CLI specification section 8.2.2 (Built-in value and reference types) has a table with the following:
System.Int32 - Signed 32-bit integer
C# specification section 8.2.1 (Predefined types) has a similar table:
int - 32-bit signed integral type
This guarantees that both System.Int32 in CLR and int in C# will always be 32-bit.
Will sizeof(testInt) ever be 8?
No, sizeof(testInt) is an error. testInt is a local variable. The sizeof operator requires a type as its argument. This will never be 8 because it will always be an error.
VS2010 compiles a c# managed integer as 4 bytes, even on a 64 bit machine.
Correct. I note that section 18.5.8 of the C# specification defines sizeof(int) as being the compile-time constant 4. That is, when you say sizeof(int) the compiler simply replaces that with 4; it is just as if you'd said "4" in the source code.
Does anyone know if/when the time will come that a standard "int" in C# will be 64 bits?
Never. Section 4.1.4 of the C# specification states that "int" is a synonym for "System.Int32".
If what you want is a "pointer-sized integer" then use IntPtr. An IntPtr changes its size on different architectures.
int is always synonymous with Int32 on all platforms.
It's very unlikely that Microsoft will change that in the future, as it would break lots of existing code that assumes int is 32-bits.
I think what you may be confused by is that int is an alias for Int32 so it will always be 4 bytes, but IntPtr is suppose to match the word size of the CPU architecture so it will be 4 bytes on a 32-bit system and 8 bytes on a 64-bit system.
According to the C# specification ECMA-334, section "11.1.4 Simple Types", the reserved word int will be aliased to System.Int32. Since this is in the specification it is very unlikely to change.
No matter whether you're using the 32-bit version or 64-bit version of the CLR, in C# an int will always mean System.Int32 and long will always mean System.Int64.
The following will always be true in C#:
sbyte signed 8 bits, 1 byte
byte unsigned 8 bits, 1 byte
short signed 16 bits, 2 bytes
ushort unsigned 16 bits, 2 bytes
int signed 32 bits, 4 bytes
uint unsigned 32 bits, 4 bytes
long signed 64 bits, 8 bytes
ulong unsigned 64 bits, 8 bytes
An integer literal is just a sequence of digits (eg 314159) without any of these explicit types. C# assigns it the first type in the sequence (int, uint, long, ulong) in which it fits. This seems to have been slightly muddled in at least one of the responses above.
Weirdly the unary minus operator (minus sign) showing up before a string of digits does not reduce the choice to (int, long). The literal is always positive; the minus sign really is an operator. So presumably -314159 is exactly the same thing as -((int)314159). Except apparently there's a special case to get -2147483648 straight into an int; otherwise it'd be -((uint)2147483648). Which I presume does something unpleasant.
Somehow it seems safe to predict that C# (and friends) will never bother with "squishy name" types for >=128 bit integers. We'll get nice support for arbitrarily large integers and super-precise support for UInt128, UInt256, etc. as soon as processors support doing math that wide, and hardly ever use any of it. 64-bit address spaces are really big. If they're ever too small it'll be for some esoteric reason like ASLR or a more efficient MapReduce or something.
Yes, as Jon said, and unlike the 'C/C++ world', Java and C# aren't dependent on the system they're running on. They have strictly defined lengths for byte/short/int/long and single/double precision floats, equal on every system.
int without suffix can be either 32bit or 64bit, it depends on the value it represents.
as defined in MSDN:
When an integer literal has no suffix, its type is the first of these types in which its value can be represented: int, uint, long, ulong.
Here is the address:
https://msdn.microsoft.com/en-us/library/5kzh1b5w.aspx

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