I don't understand why C# considers the literal 0xFFFFFFFF as a uint when it also represents -1 for int types.
The following is code was entered into the Immediate Window shown here with the output:
int i = -1;
-1
string s = i.ToString("x");
"ffffffff"
int j = Convert.ToInt32(s, 16);
-1
int k = 0xFFFFFFFF;
Cannot implicitly convert type 'uint' to 'int'. An explicit conversion exists (are you missing a cast?)
int l = Convert.ToInt32(0xFFFFFFFF);
OverflowException was unhandled: Value was either too large or too small for an Int32.
Why can the string hex number be converted without problems but the literal only be converted using unchecked?
Why is 0xFFFFFFFF a uint when it represents -1?
Because you're not writing the bit pattern when you write
i = 0xFFFFFFFF;
you're writing a number by C#'s rules for integer literals. With C#'s integer literals, to write a negative number we write a - followed by the magnitude of the number (e.g., -1), not the bit pattern for what we want. It's really good that we aren't expected to write the bit pattern, it would make it really awkward to write negative numbers. When I want -3, I don't want to have to write 0xFFFFFFFD. :-) And I really don't want to have to vary the number of leading Fs based on the size of the type (0xFFFFFFFFFFFFFFFD for a long -3).
The rule for choosing the type of the literal is covered by the above link by saying:
If the literal has no suffix, it has the first of these types in which its value can be represented: int, uint, long, ulong.
0xFFFFFFFF doesn't fit in an int, which has a maximum positive value of 0x7FFFFFFF, so the next in the list is uint, which it does fit in.
0xffffffff is 4294967295 is an UInt32 that just happens to have a bit pattern equal to the Int32 -1 due to the way negative numbers are represented on computers. Just because they have the same bit pattern, that doesn't mean 4294967295 = -1. They're completely different numbers so of course you can't just trivially convert between the two. You can force the reintepretation of the bit pattern by using an explicit cast to int: (int)0xffffffff.
The C# docs say that the compiler will try to fit the number you provide in the smallest type that can fit it. That doc is a bit old, but it applies still. It always assumes that the number is positive.
As a fallback you can always coerce the type.
The C# language rules state that 0xFFFFFFFF is an unsigned literal.
A C# signed int is 2's complement type. That scheme uses 0xFFFFFFFF to represent -1. (2's complement is a clever scheme since it doesn't have a signed zero).
For an unsigned int, 0xFFFFFFFF is the largest value it can take, and due to its size, it can't be converted to a signed int.
Related
I am wondering how you take a number (for example 9), convert it to a 32 int (00000000000000000000000000001001), then invert or flip every bit (11111111111111111111111111110110) so that the zeroes become ones and the ones become zeroes.
I know how to do that by replacing the numbers in a string, but I need to know how to do that with binary operators on a binary number.
I think you have to use this operator, "~", but it just gives me a negative number when I use it on a value.
That is doing the correct functionality. The int data type within C# uses signed integers, so 11111111111111111111111111110110 is in fact a negative number.
As Marc pointed out, if you want to use unsigned values declare your number as a uint.
If you look at the decimal version of your number then its a negative number.
If you declare it as a unsigned int then its a positive one.
But this doesnt matter, binary it will always be 11111111111111111111111111110110.
Try this:
int number = 9;
Console.WriteLine(Convert.ToString(number, 2)); //Gives you 1001
number = ~number; //Invert all bits
Console.WriteLine(Convert.ToString(number, 2));
//Gives you your wanted result: 11111111111111111111111111110110
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.
This is the code.
CODE 1: cmd.Parameters.Add("#RateCenterID", OleDbType.Integer).Value = Convert.ToInt32(ratecenterid.Text);
CODE 2: cmd.Parameters.Add("#QuantityThreshold", OleDbType.Integer).Value = Convert.ToIn32(quantityThreshold.Text);
I get the following error in the CODE 2 but not in CODE 1
Error: Value was either too large or too small for an Int32.
kindly help me
Regards,
Arjun
Try entering a smaller or larger value in the quantityThreshold text box.
Update
Okay, reading your comment response to Saeed Amiri, 4036953909 is too large for an int - it's range is -2147483648 to 2147483647
You could use a uint which has a range 0 to 4294967295
Convert.ToUInt32(ratecenterid.Text)
Although you will likely need to change the parameter type to OleDbType.UnsignedInt as well.
From MSDN - Int32 Structure:
Int32 is an immutable value type that represents signed integers with values that range from negative 2,147,483,648 (which is represented by the Int32.MinValue constant) through positive 2,147,483,647 (which is represented by the Int32.MaxValue constant. The .NET Framework also includes an unsigned 32-bit integer value type, UInt32, which represents values that range from 0 to 4,294,967,295.
You are trying to use a value that is not possible to represent as an Int32.
Not sure why your title is about Int16 as your code and error indicate you are using Int32 - regardless - Int16 has an much smaller range than Int32:
The Int16 value type represents signed integers with values ranging from negative 32768 through positive 32767.
You need to either use UInt32 or long.
sample value for "QuantityThreshold = 4036953909"
Right. That wouldn't fit into an Int16 (as per your title), which has a maximum value of 32767. That wouldn't even fit into an Int32 (as per your code) (max value 2,147,483,647). It would fit into a UInt32, and would easily fit into an Int64 - perhaps the latter is what you're really looking for?
(You should also clarify your qusetion - are you using Int32 or Int16?)
Your current value is within UInt32 scope, and OleDbType.Integer size is 4 byte and is mapped to Int32, if you want use UInt32 you should use UnsignedInt as your parameter type, and use UInt32 conversion:
cmd.Parameters.Add("#QuantityThreshold", OleDbType.UnsignedInt).Value =
Convert.ToUIn32(quantityThreshold.Text)
Int32 has a range of -2147483648 to 2147483647. You could use a larger type e.g., uint (0 to 4294967295) or long (-9.2233E+18 to 9.2233E+18).
Ref: Error: Value was either too large or too small for an Int32
That number is way to big for an int32. You should just store it as a
string.(varchar)
You got the why in the other answers but I didn't see anyone mentioning ToInt64. Simply change the ToInt32 to ToInt64 and also make sure that the datatype of #QuantityThreshold can handle it.
The Int64 value type represents integers with values ranging from negative 9,223,372,036,854,775,808 through positive 9,223,372,036,854,775,807.
To fix your problem you will have to put the validation on your textbox - quantityThreshold, it should have the value with in the range of Int32. or use a larger type.
I know about int and long (32-bit and 64-bit numbers), but what are uint and ulong?
The primitive data types prefixed with "u" are unsigned versions with the same bit sizes. Effectively, this means they cannot store negative numbers, but on the other hand they can store positive numbers twice as large as their signed counterparts. The signed counterparts do not have "u" prefixed.
The limits for int (32 bit) are:
int: –2147483648 to 2147483647
uint: 0 to 4294967295
And for long (64 bit):
long: -9223372036854775808 to 9223372036854775807
ulong: 0 to 18446744073709551615
uint and ulong are the unsigned versions of int and long. That means they can't be negative. Instead they have a larger maximum value.
Type Min Max CLS-compliant
int -2,147,483,648 2,147,483,647 Yes
uint 0 4,294,967,295 No
long –9,223,372,036,854,775,808 9,223,372,036,854,775,807 Yes
ulong 0 18,446,744,073,709,551,615 No
To write a literal unsigned int in your source code you can use the suffix u or U for example 123U.
You should not use uint and ulong in your public interface if you wish to be CLS-Compliant.
Read the documentation for more information:
int
uint
long
ulong
By the way, there is also short and ushort and byte and sbyte.
The difference is that the uint and ulong are unsigned data types, meaning the range is different: They do not accept negative values:
int range: -2,147,483,648 to 2,147,483,647
uint range: 0 to 4,294,967,295
long range: –9,223,372,036,854,775,808 to 9,223,372,036,854,775,807
ulong range: 0 to 18,446,744,073,709,551,615
u means unsigned, so ulong is a large number without sign. You can store a bigger value in ulong than long, but no negative numbers allowed.
A long value is stored in 64-bit,with its first digit to show if it's a positive/negative number. while ulong is also 64-bit, with all 64 bit to store the number. so the maximum of ulong is 2(64)-1, while long is 2(63)-1.
Based on the other answers here and a little review you can understand it this way: unsigned is in reference to the assignment of a negative or positive explicit assignment (think the "-" in -1) and the inability to have negative versions of said numbers.
And because of this capacity on the negative end being removed as an option they instead allocated that capacity to the positive end hence the doubling of the positive valuation's maximum value. So instead of the bit range being split along positive and negative valuations, they are instead for ushort, uint, along, etc allocated to the positive end of the valuation.
It's been a while since I C++'d but these answers are off a bit.
As far as the size goes, 'int' isn't anything. It's a notional value of a standard integer; assumed to be fast for purposes of things like iteration. It doesn't have a preset size.
So, the answers are correct with respect to the differences between int and uint, but are incorrect when they talk about "how large they are" or what their range is. That size is undefined, or more accurately, it will change with the compiler and platform.
It's never polite to discuss the size of your bits in public.
When you compile a program, int does have a size, as you've taken the abstract C/C++ and turned it into concrete machine code.
So, TODAY, practically speaking with most common compilers, they are correct. But do not assume this.
Specifically: if you're writing a 32 bit program, int will be one thing, 64 bit, it can be different, and 16 bit is different. I've gone through all three and briefly looked at 6502 shudder
A brief google search shows this:
https://www.tutorialspoint.com/cprogramming/c_data_types.htm
This is also good info:
https://docs.oracle.com/cd/E19620-01/805-3024/lp64-1/index.html
use int if you really don't care how large your bits are; it can change.
Use size_t and ssize_t if you want to know how large something is.
If you're reading or writing binary data, don't use int. Use a (usually platform/source dependent) specific keyword. WinSDK has plenty of good, maintainable examples of this. Other platforms do too.
I've spent a LOT of time going through code from people that "SMH" at the idea that this is all just academic/pedantic. These ate the people that write unmaintainable code. Sure, it's easy to use type 'int' and use it without all the extra darn typing. It's a lot of work to figure out what they really meant, and a bit mind-numbing.
It's fragile coding when you mix int and assume sizes.
use int and uint when you just want a fast integer and don't care about the range (other than signed/unsigned).
This is a follow up question. So, Java store's integers in two's-complements and you can do the following:
int ALPHA_MASK = 0xff000000;
In C# this requires the use of an unsigned integer, uint, because it interprets this to be 4278190080 instead of -16777216.
My question, how do declare negative values in hexadecimal notation in c#, and how exactly are integers represented internally? What are the differences to Java here?
C# (rather, .NET) also uses the two's complement, but it supports both signed and unsigned types (which Java doesn't). A bit mask is more naturally an unsigned thing - why should one bit be different than all the other bits?
In this specific case, it is safe to use an unchecked cast:
int ALPHA_MASK = unchecked((int)0xFF000000);
To "directly" represent this number as a signed value, you write
int ALPHA_MASK = -0x1000000; // == -16777216
Hexadecimal is not (or should not) be any different from decimal: to represent a negative number, you need to write a negative sign, followed by the digits representing the absolute value.
Well, you can use an unchecked block and a cast:
unchecked
{
int ALPHA_MASK = (int)0xff000000;
}
or
int ALPHA_MASK = unchecked((int)0xff000000);
Not terribly convenient, though... perhaps just use a literal integer?
And just to add insult to injury, this will work too:
-0x7F000000