Now I know that converting a int to hex is simple but I have an issue here.
I have an int that I want to convert it to hex and then add another hex to it.
The simple solution is int.Tostring("X") but after my int is turned to hex, it is also turned to string so I can't add anything to it until it is turned back to int again.
So my question is; is there a way to turn a int to hex and avoid having turned it to string as well. I mean a quick way such as int.Tostring("X") but without the int being turned to string.
I mean a quick way such as int.Tostring("X") but without the int being
turned to string.
No.
Look at this way. What is the difference between those?
var i = 10;
var i = 0xA;
As a value, they are exactly same. As a representation, first one is decimal notation and the second one is hexadecimal notation. The X you use hexadecimal format specifier which generates hexadecimal notation of that numeric value.
Be aware, you can parse this hexadecimal notation string to integer anytime you want.
C# convert integer to hex and back again
There is no need to convert. Number ten is ten, write it in binary or hex, yes their representation will differ depending in which base you write them but value is same. So just add another integer to your integer - and convert the final result to hex string when you need it.
Take example. Assume you have
int x = 10 + 10; // answer is 20 or 0x14 in Hex.
Now, if you added
int x = 0x0A + 0x0A; // x == 0x14
Result would still be 0x14. See?
Numeric 10 and 0x0A have same value just they are written in different base.
Hexadecimal string although is a different beast.
In above case that could be "0x14".
For computer this would be stored as: '0', 'x', '1', '4' - four separate characters (or bytes representing these characters in some encoding). While in case with integers, it is stored as single integer (encoded in binary form).
I guess you missing the point what is HEX and what is INT. They both represent an numbers. 1, 2, 3, 4, etc.. numbers. There's a way to look at numbers: as natural numbers: INT and hexadecimal - but at the end those are same numbers. For example if you have number: 5 + 5 = 10 (int) and A (as hex) but it the same number. Just view on them are different
Hex is just a way to represent number. The same statment is true for decimal number system and binary although with exception of some custom made numbers (BigNums etd) everything will be stored as binary as long as its integer (by integer i mean not floating point number). What would you really like to do is probably performing calculations on integers and then printing them as a Hex which have been already described in this topic C# convert integer to hex and back again
The short answer: no, and there is no need.
The integer One Hundred and seventy nine (179) is B3 in hex, 179 in base-10, 10110011 in base-2 and 20122 in base-3. The base of the number doesn't change the value of it. B3, 17, 10110011, and 20122 are all the same number, they are just represented different. So it doesn't matter what base they are in as long as you do you mathematical operations on numbers in the same base it doesn't matter what the base is.
So in your case with Hex numbers, they can contain characters such as 'A','B', 'C', and so on. So when you get a value in hex if it is a number that will contain a letter in its hex representation it will have to be a string as letters are not ints. To do what you want, it would be best to convert both numbers to regular ints and then do math and convert to Hex after. The reason for this is that if you want to be able to add (or whatever operation) with them looking like hex you are going to to need to change the behavior of the desired operator on string which is a hassle.
Related
How to convert hexadecimal string to number in C#?
I tried as below but it is giving negative value:
var dec1 = long.Parse("F0A6AFE69D2271E7", System.Globalization.NumberStyles.HexNumber);
// Result: dec1 = -1106003253459258905
While in Javascript it works fine as below:
var dec2 = parseInt("F0A6AFE69D2271E7", 16);
// Result: dec2 = 17340740820250292000
The number you're parsing is outside the range of long - long.MaxValue is 0x7FFFFFFFFFFFFFFF, and your value is 0xF0A6AFE69D2271E7.
Use ulong.Parse instead and it should be fine.
I suspect it's "working" in JavaScript because (at the time of writing) all JavaScript numbers are 64-bit floating point values, so have a huge range - but less precision, which is why a value which is clearly odd (last hex digit 7) is giving an even result.
See https://en.wikipedia.org/wiki/Two%27s_complement to understand why your number was interpreted as negative number. Presicely speaking, it's not "out of range" problem but, just your hex representation gives exact negative number. "long" type is signed integer and cannot hold full-64bit positive number since its MSB is kept for presenting negative numbers. Try using "ulong" instead.
I have a line in my function that calculates the sum of two digits.
I get the sum with this syntax:
sum += get2DigitSum((acctNumber[0] - '0') * 2);
which multiplys the number on index 0 with 2.
public static int get2DigitSum(int num)
{
return (num / 10) + (num % 10);
Lets say we have number 9 on index 0. If i have acctNumber[0] - '0' it passes the 9 into the other function.But if I don't have the - '0' after the acctNumber[0] it passes 12. I don't understand why I get wrong result if I don't use - '0'.
The text "0" and the number 0 are not at all equal to a computer.
The character '0' has in fact the ASCII number 48 (or 0x30 in hex), so to convert the character '0' into the number 0 you need to subtract 48 - in C and most languages based on it, this can be written as subtracting the character '0', which has the numerical value 48.
The beauty is, that the character '1' has the ASCII number 49, so subtracting the number 48 (or the character '0') gives 49-48=1 and so on.
So the important part is: Computers are not only sensitve to data (patterns of bits in some part of the machine), but also to the interpretation of this data - in your case interpreting it as a text and interpreting ist as a number is not the same, but gives a difference of 48, which you need to get rid of by a subtraction.
Because you are providing acctNumber[0] to get2DigitSum.
get2DigitSum accepts an integer, but acctNumber[0] is not an integer, it holds an char which represents a character with an integer value.
Therefore, you need to subtract the '0' to get the integer.
'0' to '9' have ASCII values of 48 to 57.
When you subtract two char values, actually there ASCII values get subtracted. That's why, you need to subtract '0'
Internally all Characters are represented as numbers. Numbers that then get converted into nice pictograms during display only.
Now the digits 0-9 are ASCII codes 48-57. Basically they are offset by +48. Past 57 you find the english alphabet in small and then large. And before that various operators and even a bunch of unprintable characters.
Normally you would not be doing this kind of math at all. You would feed the whole string into a Parse() or TryParse() function and then work with the parsed numbers. There are a few cases where you would not do that and isntead go for "math with Characters":
you did not know about Parse and integers when you made it
you want to support arbitary sized numbers in your calculations. This is a common beginner approach (the proper way is BigInteger).
You might be doing stuff like sorting mixed letter/number strings by the fully interpreted number (so 01 would come before 10). The same way windows sorts files with numbers in them.
You do not have a prewritten parse function. Like I did back when I started learning in C++ back in 2000.
I want to be able to convert a
varchar > decimal > long
when dbreader[11] holds a 3319419,13 as varchar
For example:
decimal d = Convert.ToDecimal(dbreader[11], CultureInfo.InvariantCulture);
I get
d = 3319419.13M
When i convert this to long object I end up with
long i = 3319419
I want to end up with 3319419,1
It never will, I'm afraid. long is a synonym for System.Int64 - a 64 bit integer value so it is unable to store the precision you are asking it to. See here
You cannot end up with 3319419,1 in a long, because long is integer.
Try using a double instead.
EDIT: Now, double happens to be a floating-point number. This means that the number of decimal digits that this type is capable of representing is not fixed, but variable. So, if you want to see a specific number of decimal digits, then this is entirely a matter of display. So, in order to see your double variable with only one digit, convert it to string as follows: String.Format( "{0:00.0}", my_double_variable );
A long is an 8 byte integer (whole number). If you require more precision, use a double (8 byte decimal)
Reality check.
3319419,1
A long is incapable of keeping partial numbers. 1, 2, 3 - not 1, 1.1, 1.2. This is teh definition of a long.
SO, you CAN NOT keep fractional values in a long, regardless what you want. Want to keep fractional values - use a data type that can.
A long does not have any decimals. It is an integer type.
You might want to use double, or keep it as a decimal.
In C#, is there a way to convert an int value to a hex value without using the .ToString("X") method?
Your question is plain wrong (no offense intended). A number has one single value. Hex, Decimal, Binary, Octal, etc. are just different representations of one same integral number. Int32 is agnostic when it comes to what representation you choose to write it with.
So when you ask:
is there a way to convert an int value to a hex value
you are asking something thast doesn't make sense. A valid question would be: is there anyway to write a integer in hexadecimal representation that doesn't involve using .ToString("X")?
The answer is not really. Someway or the other (directly or not by you), .ToString("X") or some other flavor of ToString() will be called to correctly format the string representing the value.
And when you think of hexadecimal as a respresentation (a formatted string) of a given number, then .ToString() does make sense.
Use Convert.ToString( intValue, 16 );
It can be used to convert between any common numeric base, i.e., binary, octal, decimal and hexadecimal.
I have a byte array represented by hex values, these are time durations. The data could be converted to integer values and multiplied by a constant to get the timings. The decoding of the data will be saved to a file as a series of hex strings. What would be an efficient way of manipulating hex values?
I was looking at performance issues when dealing with data formats, as I have to work with more than one format at different stages (calculations, data display, etc.). Most examples show the conversion from byte[] to hex string ("1A 3C D4"), and viceversa, but I was looking for an alternative, which is to convert to Int16 and use char[] array.
You don't have a byte array representing hex values. You have a byte array representing numbers. The base you represent a number in is only relevant when you're representing it.
To put it a different way: if you thought of your byte array as representing decimal integers instead, how do you imagine it would be different? Is my height different if I represent it in feet and inches instead of metres?
Now, if you're trying to represent 16-bit numbers, I'd suggest that using a byte array is a bad idea. Use a ushort[] or short[] instead, as those are 16-bit values. If you're having trouble getting the data into an array like that, please give details... likewise if you have any other problems with the manipulation. Just be aware that until you're writing the data out as text, there's really no such concept as which base it's in, as far as the computer is concerned.
(Note that this is different for floating point values, where the data really would be different between a decimal and a double, for example... there, the base of representation is part of the data format. It's not for integers. Alternatively, you can think of all integers as just being binary until you decide to format them as text...)
From MSDN:
The hexadecimal ("X") format specifier
converts a number to a string of
hexadecimal digits. The case of the
format specifier indicates whether to
use uppercase or lowercase characters
for hexadecimal digits that are
greater than 9. For example, use "X"
to produce "ABCDEF", and "x" to
produce "abcdef". This format is
supported only for integral types.
The precision specifier indicates the
minimum number of digits desired in
the resulting string. If required, the
number is padded with zeros to its
left to produce the number of digits
given by the precision specifier.
byte x = 60;
string hex = String.Format("0x{0:X4}", x);
Console.WriteLine(hex); // prints "0x003C"