This method convert a double value into its bit representation formatted in hexadecimal format. Why is specified that the converted value is a signed int64 ? Which is the difference with unsigned int64 ? The binary representation of the double is unique.
First off, that method does not convert it to hexadecimal, it converts the bits to be interpreted as an Int64, which you could then convert to a hexadecimal string if you chose, using ToString("X") or some such.
There is no difference between a signed Int64 and unsigned UInt64 under the hood-- the only difference is how the 64-bit word is interpreted by your program.
A 64-bit double is also a 64-bit word, under the hood-- but some processors, like intel x86 family of processors, support an 80-bit double precision floating point, not just a 64-bit double. That method is simply allowing C# to interpret the bits for a 64-bit double (or convert an 80-bit double to a 64-bit double, if that's what it has to work with) as if it were a 64-bit signed int. If you want to interpret that as a 64-bit unsigned int, just cast it (using unchecked to prevent c# from freaking out):
ulong bitsForDouble;
unchecked {
bitsForDouble = (ulong)BitConverter.DoubleToInt64Bits(someDoubleValue);
}
It doesn't make sense to interpret the bits of a double as either an Int64 or a UInt64-- they just wanted to give us a way to represent a double in a 64-bit value for whatever bit manipulation needs we might have.
Related
I am reading values from a S7-300 PLC with my c# code. When the values are in INT format there is no problem. But there are some 32 bit memory areas (Double Words) that are encoded in IEEE 754 Floating-Point standart. (First bit is sign bit, the next 8 bits exponent, and the remaining 23 bits Mantissa)
I can read out this memory areas from the PLC only as Int32 (As they were integer).
How can I convert this as integer read value to a single Real value in C# with respeect of the IEEE 754 Floating-Point encoding in the double word?
It worked just as wanted with Eldar's answer.
If you read a 32 bit float value as bit, then just convert it like this:
Thanks again to Eldar :-)
var finalSingle= BitConverter.ToSingle(BitConverter.GetBytes(s7Int))
In C#:
0x80000000==2147483648 //outputs True
In VB.NET:
&H80000000=2147483648 'outputs False
How is this possible?
This is related to the history behind the languages.
C# always supported unsigned integers. The value you use are too large for int so the compiler picks the next type that can correctly represent the value. Which is uint for both.
VB.NET didn't acquire unsigned integer support until version 8 (.NET 2.0). So traditionally, the compiler was forced to pick Long as the type for the 2147483648 literal. The rule was however different for the hexadecimal literal, it traditionally supported specifying the bit pattern of a negative value (see section 2.4.2 in the language spec). So &H80000000 is a literal of type Integer with the value -2147483648 and 2147483648 is a Long. Thus the mismatch.
If you think VB.NET is a quirky language then I'd invite you to read this post :)
The VB version should be:
&H80000000L=2147483648
Without the 'long' specifier ('L'), VB will try to interpret &H8000000 as an integer. If you force it to consider this as a long type, then you'll get the same result.
&H80000000UI will also work - actually this is the type (UInt32) that C# regards the literal as.
This happens because the type of the hexadecimal number is UInt32 in C# and Int32 in VB.NET.
The binary representation of the hexadecimal number is:
10000000000000000000000000000000
Both UInt32 and Int32 take 32 bits, but because Int32 is signed, the first bit is considered a sign to indicate whether the number is negative or not: 0 for positive, 1 for negative. To convert a negative binary number to decimal, do this:
Invert the bits. You get 01111111111111111111111111111111.
Convert this to decimal. You get 2147483647.
Add 1 to this number. You get 2147483648.
Make this negative. You get -2147483648, which is equal to &H80000000 in VB.NET.
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.
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