I have a C struct that is defined as follows:
typedef struct {
unsigned long int a;
} TEST;
I want to create a C# equivalent of this struct?
Any help? What is confusing me is that "unsigned long int" is at least 32-bit, what does that mean, it's either 32-bit, 64-bit or 16-bit, right?
You want an uint or ulong depending on what an int or long was on your native C platform:
C# uint is 32 bits
C# ulong is 64 bits
The at least and platform dependency is a necessary concern in C because it is actually translated into machine code and C was developed for many architectures with varying word sizes. C# on the contrary is defined against a virtual machine (exactly like Java or Javascript) and thus can abstract the hardware's word size in favor of one defined by the language's standard VM (the CLR in C#). Differences between the VM and harware word size are taken care of by the VM and hidden to the hosted code.
Related
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
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
What is the actual difference between a long and an int in C#? I understand that in C/C++ long would be 64bit on some 64bit platforms(depending on OS of course) but in C# it's all running in the .NET runtime, so is there an actual distinction?
Another question: can an int hold a long(by cast) without losing data on all platforms?
An int (aka System.Int32 within the runtime) is always a signed 32 bit integer on any platform, a long (aka System.Int64) is always a signed 64 bit integer on any platform. So you can't cast from a long with a value above Int32.MaxValue or below Int32.MinValue without losing data.
int in C#=> System.Int32=>from -2,147,483,648 to 2,147,483,647.
long in C#=> System.Int64 =>from -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807
If your long data exceeds the range of int, and you use Convert.ToInt32 then it will throw OverflowException, if you use explicit cast then the result would be unexpected.
int is 32 bits in .NET. long is 64-bits. That is guaranteed. So, no, an int can't hold a long without losing data.
There's a type whose size changes depending on the platform you're running on, which is IntPtr (and UIntPtr). This could be 32-bits or 64-bits.
Sure is a difference - In C#, a long is a 64 bit signed integer, an int is a 32 bit signed integer, and that's the way it always will always be.
So in C#, a long can hold an int, but an int cannot hold a long.
C/C++ that question is platform dependent.
In C#, an int is a System.Int32 and a long is a System.Int64; the former is 32-bits and the later 64-bits.
C++ only provides vague guarantees about the size of int/long, in comparison (you can dig through the C++ standard for the exact, gory, details).
I think an int is a 32-bit integer, while a long is a 64-bit integer.
I am using C# in Mono and I'm trying to use pinvoke to call a Linux shared library.
The c# call is defined as:
[DllImport("libaiousb")]
extern static ulong AIOUSB_Init();
The Linux function is defined as follows:
unsigned long AIOUSB_Init() {
return(0);
}
The compile command for the Linux code is:
gcc -ggdb -std=gnu99 -D_GNU_SOURCE -c -Wall -pthread -fPIC
-I/usr/include/libusb-1.0 AIOUSB_Core.c -o AIOUSB_Core.dbg.o
I can call the function ok but the return result is bonkers. It should be 0 but I'm getting some huge mangled number.
I've put printf's in the Linux code just before the function value is returned and it is correct.
One thing I have noticed that is a little weird is that the printf should occur before the function returns. However, I see the function return to C# and then the c# prints the return result and finally the printf result is displayed.
From here:
An unsigned long can hold all the values between 0 and ULONG_MAX inclusive. ULONG_MAX must be at least 4294967295. The long types must contain at least 32 bits to hold the required range of values.
For this reason a C unsigned long is usually translated to a .NET UInt32:
[DllImport("libaiousb")]
extern static uint AIOUSB_Init();
You're probably running that on a system where C unsigned long is 32-bits. C# unsigned long is 64 bits. If you want to make sure the return value is a 64-bits unsigned long, include stdint.h and return an uint64_t from AIOUSB_Init().
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