Break up 32-bit hex value into 4 bytes [QB64] - memory

I would want to ask, how do you break up a 32-bit hex (for example: CEED6644) into 4 bytes (var1 = CE, var2 = ED, var3 = 66, var4 = 44). In QB64 or QBasic. I would use this to store several data bytes into one array address.
Something like this:
DIM Array(&HFFFF&) AS _UNSIGNED LONG
Array(&HAA00&) = &HCEED6644&
addr = &HAA00&
SUB PrintChar
SHARED addr
IF var1 = &HAA& THEN PRINT "A"
IF var1 = &HBB& THEN PRINT "B"
IF var1 = &HCC& THEN PRINT "C"
IF var1 = &HDD& THEN PRINT "D"
IF var1 = &HEE& THEN PRINT "E"
IF var1 = &HFF& THEN PRINT "F"
IF var1 = &H00& THEN PRINT "G"
IF var1 = &H11& THEN PRINT "H"
And so on...

You could use integer division (\) and bitwise AND (AND) to accomplish this.
DIM x(0 TO 3) AS _UNSIGNED _BYTE
a& = &HCEED6644&
x(0) = (a& AND &HFF000000&) \ 2^24
x(1) = (a& AND &H00FF0000&) \ 2^16
x(2) = (a& AND &H0000FF00&) \ 2^8
x(3) = a& AND &HFF&
PRINT HEX$(x(0)); HEX$(x(1)); HEX$(x(2)); HEX$(x(3))
Note that you could alternatively use a generic RShift~& function instead of raw integer division since what you're really doing is shifting bits:
x(0) = RShift~&(a& AND &HFF000000&, 18)
...
FUNCTION RShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED BYTE)
' Raise illegal function call if the shift count is greater than the width of the type.
' If shiftCount is not _UNSIGNED, then you must also check that it isn't less than 0.
IF shiftCount > 32 THEN ERROR 5
RShift~& = value / 2^shiftCount
END FUNCTION
Building upon that, you might create another function:
FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE)
'position must be in the range [0, 3].
IF (position AND 3) <> position THEN ERROR 5
ByteAt~%% = RShift~&(value AND LShift~&(&HFF&, 8*position), 8*position)
END FUNCTION
Note that an LShift~& function was used that shifts bits to the left (multiplication by a power of 2). A potentially better alternative would be to perform the right-shift first and just mask the lower 8 bits, eliminating the need for LShift~&:
FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE)
'position must be in the range [0, 3].
IF (position AND 3) <> position THEN ERROR 5
ByteAt~%% = RShift~&(value, 8*position) AND 255
END FUNCTION
Incidentally, another QB-like implementation known as FreeBASIC has an actual SHR operator, used like MOD or AND, to perform a shift operation directly instead of using division, which is potentially faster.
You could also use QB64's DECLARE LIBRARY facility to create functions in C++ that will perform the shift operations:
/*
* Place in a separate "shift.h" file or something.
*/
unsigned int LShift (unsigned int n, unsigned char count)
{
return n << count;
}
unsigned int RShift (unsigned int n, unsigned char count)
{
return n >> count;
}
Here's the full corresponding QB64 code:
DECLARE LIBRARY "shift"
FUNCTION LShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED _BYTE)
FUNCTION RShift~& (value AS _UNSIGNED LONG, shiftCount AS _UNSIGNED _BYTE)
END DECLARE
x(0) = ByteAt~%%(a&, 0)
x(1) = ByteAt~%%(a&, 1)
x(2) = ByteAt~%%(a&, 2)
x(3) = ByteAt~%%(a&, 3)
END
FUNCTION ByteAt~%% (value AS _UNSIGNED LONG, position AS _UNSIGNED BYTE)
'position must be in the range [0, 3].
IF (position AND 3) <> position THEN ERROR 5
ByteAt~%% = RShift~&(value, 8*position) AND 255
END FUNCTION
If QB64 had a documented API, it might be possible to raise a QB64 error from the C++ code when the shift count is too high, rather than relying on the behavior of C++ to essentially ignore shift counts that are too high. Unfortunately, this isn't the case, and it might actually cause more problems than it's worth.

This snip gets the byte pairs of a hexidecimal value:
DIM Value AS _UNSIGNED LONG
Value = &HCEED6644&
S$ = RIGHT$("00000000" + HEX$(Value), 8)
PRINT "Byte#1: "; MID$(S$, 1, 2)
PRINT "Byte#2: "; MID$(S$, 3, 2)
PRINT "Byte#3: "; MID$(S$, 5, 2)
PRINT "Byte#4: "; MID$(S$, 7, 2)

Related

Convert bytes to signed integers in lua 5.1.5

I'm looking for how to turn bytes into a signed int using lua 5.1.5, so far I've only been able to find solutions for lua 5.2 onward, and they are not backward compatible.
I have solutions for how to turn bytes into unsigned integers, like so:
payload_t.temperature=tonumber(utility.hex2str(string.sub(payload,32,33)),16)
First of all I'll assume that you actually have a byte string rather than a hex string given; if your string is a hex string, you can trivially convert it to a byte string using gsub:
function hex2bytes(str)
-- assert that it is indeed a string of hex digit pairs
assert(#str % 2 == 0 and not str:match"[^%x]")
return str:gsub("%x%x", function(hex) return tonumber(hex, 16) end)
end
Now, let's convert this byte string to an integer. I'll assume little endian (least significant byte first); should your string be big endian (most significant byte first) you'll have to reverse it using str:reverse() before you read it.
Reading an unsigned integer is pretty straightforward:
function bytes2uint(str)
local uint = 0
for i = 1, #str do
uint = uint + str:byte(i) * 0x100^(i-1)
end
return uint
end
I'll assume your integers are stored using Two's complement. In this case the higher 2^n values (equivalent to the first bit being set or the value being >= 2^(n-1)) the uint can take represent negative numbers, with the smallest value (2^(n-1)) representing the largest negative value (-2^(n-1)). Thus you can simply subtract the unsigned value from 2^n, the (exclusive) max value for the uint:
function bytes2int(str)
local uint = bytes2uint(str)
local max = 0x100 ^ #str
if uint >= max / 2 then
return uint - max
end
return uint
end

How to get each individual digit of a given number in Basic?

I have one program downloaded from internet and need to get each digit printed out from a three digit number. For example:
Input: 123
Expected Output:
1
2
3
I have 598
Need to Get:
5
9
8
I try using this formula but the problem is when number is with decimal function failed:
FIRST_DIGIT = (number mod 1000) / 100
SECOND_DIGIT = (number mod 100) / 10
THIRD_DIGIT = (number mod 10)
Where number is the above example so here is calulation:
FIRST_DIGIT = (598 mod 1000) / 100 = 5,98 <== FAILED...i need to get 5 but my program shows 0 because i have decimal point
SECOND_DIGIT = (598 mod 100) / 10 = 9,8 <== FAILED...i need to get 9 but my program shows 0 because i have decimal point
THIRD_DIGIT = (598 mod 10) = 8 <== CORRECT...i get from program output number 8 and this digit is correct.
So my question is is there sample or more efficient code that get each digit from number without decimal point? I don't want to use round to round nearest number because sometime it fill failed if number is larger that .5.
Thanks
The simplest solution is to use integer division (\) instead of floating point division (/).
If you replace each one of your examples with the backslash (\) instead of forward slash (/) they will return integer values.
FIRST_DIGIT = (598 mod 1000) \ 100 = 5
SECOND_DIGIT = (598 mod 100) \ 10 = 9
THIRD_DIGIT = (598 mod 10) = 8
You don't have to do any fancy integer calculations as long as you pull it apart from a string:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
PRINT MID$(X$, Z, 1)
NEXT
Then, for example, you could act upon each string element:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
NEXT
Additionally, you could tear apart the string character by character:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " ", ".", "+", "-", "E", "D"
' special char
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
I'm no expert in Basic but looks like you have to convert floating point number to Integer. A quick google search told me that you have to use Int(floating_point_number) to convert float to integer.
So
Int((number mod 100)/ 10)
should probably the one you are looking for.
And, finally, all string elements could be parsed:
INPUT X
X$ = STR$(X)
PRINT X$
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " "
' nul
CASE "E", "D"
Exponent = -1
CASE "."
Decimal = -1
CASE "+"
UnaryPlus = -1
CASE "-"
UnaryNegative = -1
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
IF Exponent THEN PRINT "There was an exponent."
IF Decimal THEN PRINT "There was a decimal."
IF UnaryPlus THEN PRINT "There was a plus sign."
IF UnaryNegative THEN PRINT "There was a negative sign."

How to set 5 bits to value 3 at offset 387 bit in byte data sequence?

I need set some bits in ByteData at position counted in bits.
How I can do this?
Eg.
var byteData = new ByteData(1024);
var bitData = new BitData(byteData);
// Offset in bits: 387
// Number of bits: 5
// Value: 3
bitData.setBits(387, 5, 3);
Yes it is quite complicated. I dont know dart, but these are the general steps you need to take. I will label each variable as a letter and also use a more complicated example to show you what happens when the bits overflow.
1. Construct the BitData object with a ByteData object (A)
2. Call setBits(offset (B), bits (C), value (D));
I will use example values of:
A: 11111111 11111111 11111111 11111111
B: 7
C: 10
D: 00000000 11111111
3. Rather than using an integer with a fixed length of bits, you could
use another ByteData object (D) containing your bits you want to write.
Also create a mask (E) containing the significant bits.
e.g.
A: 11111111 11111111 11111111 11111111
D: 00000000 11111111
E: 00000011 11111111 (2^C - 1)
4. As an extra bonus step, we can make sure the insignificant
bits are really zero by ANDing with the bitmask.
D = D & E
D 00000000 11111111
E 00000011 11111111
5. Make sure D and E contain at least one full zero byte since we want
to shift them.
D 00000000 00000000 11111111
E 00000000 00000011 11111111
6. Work out these two integer values:
F = The extra bit offset for the start byte: B mod 8 (e.g. 7)
G = The insignificant bits: size(D) - C (e.g. 14)
7. H = G-F which should not be negative here. (e.g. 14-7 = 7)
8. Shift both D and E left by H bits.
D 00000000 01111111 10000000
E 00000001 11111111 10000000
9. Work out first byte number (J) floor(B / 8) e.g. 0
10. Read the value of A at this index out and let this be K
K = 11111111 11111111 11111111
11. AND the current (K) with NOT E to set zeros for the new bits.
Then you can OR the new bits over the top.
L = (K & !E) | D
K & !E = 11111110 00000000 01111111
L = 11111110 01111111 11111111
12. Write L to the same place you read it from.
There is no BitData class, so you'll have to do some of the bit-pushing yourself.
Find the corresponding byte offset, read in some bytes, mask out the existing bits and set the new ones at the correct bit offset, then write it back.
The real complexity comes when you need to store more bits than you can read/write in a single operation.
For endianness, if you are treating the memory as a sequence of bits with arbitrary width, I'd go for little-endian. Endianness only really makes sense for full-sized (2^n-bit, n > 3) integers. A 5 bit integer as the one you are storing can't have any endianness, and a 37 bit integer also won't have any natural way of expressing an endianness.
You can try something like this code (which can definitely be optimized more):
import "dart:typed_data";
void setBitData(ByteBuffer buffer, int offset, int length, int value) {
assert(value < (1 << length));
assert(offset + length < buffer.lengthInBytes * 8);
int byteOffset = offset >> 3;
int bitOffset = offset & 7;
if (length + bitOffset <= 32) {
ByteData data = new ByteData.view(buffer);
// Can update it one read/modify/write operation.
int mask = ((1 << length) - 1) << bitOffset;
int bits = data.getUint32(byteOffset, Endianness.LITTLE_ENDIAN);
bits = (bits & ~mask) | (value << bitOffset);
data.setUint32(byteOffset, bits, Endianness.LITTLE_ENDIAN);
return;
}
// Split the value into chunks of no more than 32 bits, aligned.
do {
int bits = (length > 32 ? 32 : length) - bitOffset;
setBitData(buffer, offset, bits, value & ((1 << bits) - 1));
offset += bits;
length -= bits;
value >>= bits;
bitOffset = 0;
} while (length > 0);
}
Example use:
main() {
var b = new Uint8List(32);
setBitData(b.buffer, 3, 8, 255);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 13, 6*4, 0xffffff);
print(b.map((v)=>v.toRadixString(16)));
setBitData(b.buffer, 47, 21*4, 0xaaaaaaaaaaaaaaaaaaaaa);
print(b.map((v)=>v.toRadixString(16)));
}

Bitwise right shift >> in Objective-C

I have a variable (unsigned int) part_1.
If I do this:
NSLog(#"%u %08x", part_1, part_1); (print unsigned value, and hex value) it outputs:
2063597568 7b000000
(only first two will have values).
I want to convert this to
0000007b
So i've tried doing
unsigned int part_1b = part_1 >> 6 (and lots of variations)
But this outputs:
32243712 01ec0000
Where am i going wrong?
You want to shift by 6*4 = 24 bits, not just 6 bits. Each '0' in the hex printf represents 4 bits.
unsigned int part_1b = part_1 >> 24;
^^

How do you convert 8-bit bytes to 6-bit characters?

I have a specific requirement to convert a stream of bytes into a character encoding that happens to be 6-bits per character.
Here's an example:
Input: 0x50 0x11 0xa0
Character Table:
010100 T
000001 A
000110 F
100000 SPACE
Output: "TAF "
Logically I can understand how this works:
Taking 0x50 0x11 0xa0 and showing as binary:
01010000 00010001 10100000
Which is "TAF ".
What's the best way to do this programmatically (pseudo code or c++). Thank you!
Well, every 3 bytes, you end up with four characters. So for one thing, you need to work out what to do if the input isn't a multiple of three bytes. (Does it have padding of some kind, like base64?)
Then I'd probably take each 3 bytes in turn. In C#, which is close enough to pseudo-code for C :)
for (int i = 0; i < array.Length; i += 3)
{
// Top 6 bits of byte i
int value1 = array[i] >> 2;
// Bottom 2 bits of byte i, top 4 bits of byte i+1
int value2 = ((array[i] & 0x3) << 4) | (array[i + 1] >> 4);
// Bottom 4 bits of byte i+1, top 2 bits of byte i+2
int value3 = ((array[i + 1] & 0xf) << 2) | (array[i + 2] >> 6);
// Bottom 6 bits of byte i+2
int value4 = array[i + 2] & 0x3f;
// Now use value1...value4, e.g. putting them into a char array.
// You'll need to decode from the 6-bit number (0-63) to the character.
}
Just in case if someone is interested - another variant that extracts 6-bit numbers from the stream as soon as they appear there. That is, results can be obtained even if less then 3 bytes are currently read. Would be useful for unpadded streams.
The code saves the state of the accumulator a in variable n which stores the number of bits left in accumulator from the previous read.
int n = 0;
unsigned char a = 0;
unsigned char b = 0;
while (read_byte(&byte)) {
// save (6-n) most significant bits of input byte to proper position
// in accumulator
a |= (b >> (n + 2)) & (077 >> n);
store_6bit(a);
a = 0;
// save remaining least significant bits of input byte to proper
// position in accumulator
a |= (b << (4 - n)) & ((077 << (4 - n)) & 077);
if (n == 4) {
store_6bit(a);
a = 0;
}
n = (n + 2) % 6;
}

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