I would like to covert a decimal number (between 0 to 65536) to a hex number. Can I do it in Arduino script? Thanks
You can use sprintf to format a number as hex, e.g. something like
//lets be sure our integer is in desired range
myinteger=min(max(myinteger, 0), 65535);
//buffer big enough for 4 hex digits + terminating null
char hexbuffer[5];
sprintf(hexbuffer, "%04x", myinteger);
Related
I have this values in a file I'm working in, 0XB6B5 equals to 27500 and 0X98BA equals to "30000". Can't understand how those numbers converted to this hex values. Thanks.
I know int 27500 in hex = 0x6c6b AND int 30000 = 0x3075 but in this file values are deferent
How can I know what these bunch of hex code means?
02 00 A0 E3 1E FF 2F E1
Any convertor of these codes to decimal code like 1,2,3 etc or vice versa like deciaml code to this type of hex code?
Thanks
This is my first response in stack overflow. So here goes...
What Hex Code (a.k.a. hexadecimal) represents purely depends on its context, or what does it mean to the program or machine. It could be a string, machine code (assembly language), flags, pointers to memory, data, part of an image or whatever. And this is dependent on the processor where this code is located also.
Each 2-digit hex code is a byte and represents decimal number (0-255 or 00-FF), half of a byte or 1 digit hex code is called a nibble.
Converting Hex Code to decimal is trivial. Convert from decimal to hex, not as trivial.
There are many calculators that have this functionality built in.
0-9 => 0 – 9, A=10, B=11, C=12, D=13, E=14, F=15.
Now, if you want to convert a 2 digit number like 12 hex (i.e. 0x12 or 12h ). Here is the formula.
(16 x 1) + (1 x 2) = 18 (decimal)
A four-digit hexadecimal 4A3E =>
(4096 x 4) + (256 x 10) + (16 x 3) + (1 x 14) = 19006 (decimal)
An integer in C# is 4 bytes, so your example hex code could also represent 2 integers in C#. Or it could be simply 1 number in C# called a “long” which is 8 bytes and could represent a number between:
0 to 18,446,744,073,709,551,615 unsigned long OR
-9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 signed long
Also something to note hex code also represent characters called ASCII (pronounced a·skee) This is an internal mapping within the microprocessor and could be different. 00 is not mapped so it typically means the end of the string.
Hex codes like that could represent a binary number. You could paste "0200A0E31EFF2FE1" into a converter like this to find out that the decimal representation of that number is "144292085413916641", for example.
But, from the way that your hex codes are grouped, it appears that you're looking at binary data, rather than a single integer that's represented in hexadecimal. When hex codes are grouped in pairs, each group of two characters represents one byte. https://en.wikipedia.org/wiki/Hexadecimal#Written_representation
I am reading binary files in R and need to read 10 bytes, which must be interpreted as 4 bit unsigned integers (2 per byte, so 20 values in range 0..15 I guess).
From my understanding of the docs, this cannot done with readBin directly because the minimal length to read, 1, means 1 byte.
So I think I need to read the data as 1 byte integers and use bit-wise operations to get the 4 bit integers. I found out that the values are stored as 32 bit integers internally by R, and I found this explanation on SO that seems to describe what I want to do. So here is my attempt at an R function that follows the advice:
#' #title Interprete bits start_index to stop_index of input int8 as unsigned integer.
uint8bits <- function(int8, start_index, stop_index) {
num_bits = stop_index - start_index + 1L;
bitmask = bitwShiftL((bitwShiftL(1L, num_bits) -1L), stop_index);
return(bitwShiftR(bitwAnd(int8, bitmask), start_index));
}
However, it does not work as intended, e.g, to get the two numbers out of the read value (255 in this example), I would call the function once to extract bits 1 to 4, and once more for bits 5 to 8:
value1 = uint8bits(255L, 1, 4); # I would expect 15, but the output is 120.
value2 = uint8bits(255L, 5, 8); # I would expect 15, but the output is 0.
What am I doing wrong?
We can use the packBits function to achieve your expected behaviour:
uint8.to.uint4 <- function(int8,start_index,stop_index)
{
bits <- intToBits(int8)
out <- packBits(c(bits[start_index:stop_index],
rep(as.raw(0),32-(stop_index-start_index+1))),type="integer")
return(out)
}
uint8.to.uint4(255L,1,4)
[1] 15
We first convert the integer to a bit vector, then extract the bits you like and pad the number with 0 to achieve the 32bit internal storage length for integers (32 bits). Then we can just convert with the packBits function back to an integer
Assume this number 173250103518582539668252657343418508842, if I wanted to convert it to a hexadecimal number such that a 10 = F, 11 = E, etc. where are the breaks/how does that work?
I've done a bit of research online and I can't seem to find the answer. It's a really low-level question, I know.
6 characters in there's a 10, would that be flipped to an F or would that get missed because whatever triggers the flip in the int -> string hexadecimal conversion happens another way?
Hexadecimal is an encoding used to express binary data in base-16 where the ascending sequence is 0-9a-f (upper or lower case a-f), once character per 4-bits (4-bits has 16 possible values). Thus 2 hex characters per byte.
binary bits (msb on left) and hexadecimal:
0000 0
0001 1
0010 2
0011 3
...
1001 9
1010 a
...
1111 f
To say "10 = F, 11 = E" is not hexadecimal.
To encode the decimal number 173250103518582539668252657343418508842 convert it is a Big Integer and then hexadecimal encode the underlying bytes to hexadecimal.
or
To encode the ASCI string "173250103518582539668252657343418508842" to hexadecimal convert each letter to the underlying ASCII binary code and then encode that into hexadecimal: "313733323530313033353138353832353339363638323532363537333433343138353038383432".
See Hexadecimal and ASCII.
Aside: My first day as a programmer I had to know hex, binary and ASCII encoding, funny how things change.
I want to convert the Hexadecimal values to voltage conversion as mentioned below,
2 Byte Signed 2s Comp Binary Fraction with Binary Point to the right of the most significant bit. 1:512V scaling.
Example :
0x2A80 → 170.00 V
0xD580 → ‐170.00 V
But the 0x2A80 conversion gives me 10880 decimal value. How can i get 170.00 V from 0x2A80?
If 0x2A80 is 170.00, then that means you have 10 bits before the point and 6 bits after the point. Or in other words, you have 10880/64 == 170.
Your question seems to contain a few misconceptions:
The fact that 170.0 is a voltage is irrelevant. Numbers work the same no matter whether they are voltages, distances, or just numbers without a unit.
In most programming languages, you don't have "decimal" or "hexadecimal" values, you just have values. Decimal and hexadecimal only come in when you're dealing with text output and string. 0x2A80 is 10880, and 0xD580 is -10880.
If you happen to be programming in C:
short fixedPointNumber;
float floatingPointNumber;
scanf("%hx", &fixedPointNumber);
floatingPointNumber = fixedPointNumber / 64.0f;
printf("Converted number: %f\n", floatingPointNumber);