MIPS; getting the LSB of a hexadecimal value - hex

I understand that there are a few questions already addressing this. However, my question varies in some sort. Suppose I have to get the LSB of a value (hexadecimal) stored in a register; for e.g;
If register $t0 contains the value 0xA4, I need to obtain and store the value 4
If register $t0 contains the value 0xBF, I need to obtain and store the value F
I understand that the bitwise ANDoperation works for decimal values. Could someone please provide some assistance as to how I go about getting the LSB?
Kind regards

You can easily AND the number from which you want to extract the LSB itself like this:
0xA4 AND 0x0F
is same as (in binary)
10100100b AND 00001111
this basically means that only last four digits will be extracted from binary number and that is the LSB you want.

All the binary operations works on general purpose registers along w/ masks which are merely pure numbers (regardless of their underlying basis repr)
Even though x86 isn't MIPS you should have something like this
and EAX, 0xF

Related

How to store negative numbers in EEPROM (Arduino IDE)?

I am trying to find a straightforward way to store negative values in EEPROM, integer values ranging from -20 to 20. I have been using EEPROM.write and EEPROM.read functions to store strings one character at a time, but I am having trouble with negative numbers. I figure I only need one byte for this value.
It's just matter of number representation. You just have to use correct data types to print or use:
Version 1: int8_t data = EEPROM.read(addr);
Version 2:
byte data = EEPROM.read(addr);
Serial.print((int8_t)data);
EEPROM.write can be used directly with int8_t: EEPROM.write(int8_value);
Or, if you wan't int, put/get methods can be used for it (even for structs containing POD types only or so)

Change bit state in Julia

In Julia language how can one set, clear and reverse a single bit? I hope you won't consider this question out of scope or too broad; if so, please comment it instead of downvote it.
Having this paragraph from Julia-Lang doc:
Currently, only sizes that are multiples of 8 bits are supported.
Therefore, boolean values, although they really need just a single
bit, cannot be declared to be any smaller than eight bits.
First it's possible to take a look at the binary representation of a variable like this:
julia> bits(Int(10))
"00000000000000000000000000001010"
and secondly, one can create byte value directly using its binary form:
julia> val=0b100
0x04
julia> typeof(val)
UInt8
and lastly, the best way to change value of a bit is performing right binary operation on its byte value:
julia> val | 0b10 # set 2nd bit
0x06
julia> bits(ans)
"00000110"
julia> val & 0b11111011 # clear 3nd bit
0x00
I assume you are wanting to clear, set and check the states of specific bits in a byte.
Where:
N represents the integer in question and;
n is the number of the bit in question (i.e: the n-th bit) and;
the LSB is 1-th bit
set the n-th bit: N |= 2^(n-1)
clear the n-th bit: N &= ~(2^(n-1))
check the state of a bit by copying and shifting: (N >> (n-1)) & 1 != 0
check the state of a bit using a bit-mask: mask = 2^(n-1); N & mask != mask
reverse/toggle/invert the n-th bit, using XOR: N ⊻= 2^(n-1). The xor function may also be used.

Dereference pointers in XMM register (gather)

If I have some pointer or pointer-like values packed into an SSE or AVX register, is there any particularly efficient way to dereference them, into another such register? ("Particularly efficient" meaning "more efficient than just using memory for the values".) Is there any way to dereference them all without writing an intermediate copy of the register out to memory?
Edit for clarification: that means, assuming 32-bit pointers and SSE, to index into four arbitrary memory areas at once with the four sections of an XMM register and return four results at once to another register. Or as close to "at once" as possible. (/edit)
Edit2: thanks to PaulR's answer I guess the terminology I'm looking for is "gather", and the question therefore is "what's the best way to implement gather for systems pre-AVX2?".
I assume there isn't an instruction for this since ...well, one doesn't appear to exist as far as I can tell and anyway it doesn't seem to be what SSE is designed for at all.
("Pointer-like value" meaning something like an integer index into an array pretending to be the heap; mechanically very different but conceptually the same thing. If, say, one wanted to use 32-bit or even 16-bit values regardless of the native pointer size, to fit more values in a register.)
Two possible reason I can think of why one might want to do this:
thought it might be interesting to explore using the SSE registers for general-purpose... stuff, perhaps to have four identical 'threads' processing potentially completely unrelated/non-contiguous data, slicing through the registers "vertically" rather than "horizontally" (i.e. instead of the way they were designed to be used).
to build something like romcc if for some reason (probably not a good one), one didn't want to write anything to memory, and therefore would need more register storage.
This might sound like an XY problem, but it isn't, it's just curiosity/stupidity. I'll go looking for nails once I have my hammer.
The question is not entirely clear, but if you want to dereference vector register elements then the only instructions which might help you here are AVX2's gathered loads, e.g. _mm256_i32gather_epi32 et al. See the AVX2 section of the Intel Intrinsics Guide.
SYNOPSIS
__m256i _mm256_i32gather_epi32 (int const* base_addr, __m256i vindex, const int scale)
#include "immintrin.h"
Instruction: vpgatherdd ymm, vm32x, ymm
CPUID Flag : AVX2
DESCRIPTION
Gather 32-bit integers from memory using 32-bit indices. 32-bit elements are loaded from addresses starting at base_addr and offset by each 32-bit element in vindex (each index is scaled by the factor in scale). Gathered elements are merged into dst. scale should be 1, 2, 4 or 8.
OPERATION
FOR j := 0 to 7
i := j*32
dst[i+31:i] := MEM[base_addr + SignExtend(vindex[i+31:i])*scale]
ENDFOR
dst[MAX:256] := 0
So if I understood this correctly, your title is misleading and you really want to:
index into the concatenation of all XMM registers
with an index held in a part of an XMM register
Right?
That's hard. And a little weird, but I'm OK with that.
Assuming crazy tricks are allowed, I propose self-modifying code: (not tested)
pextrb eax, xmm?, ? // question marks are the position of the pointer
mov edx, eax
shr eax, 1
and eax, 0x38
add eax, 0xC0 // C0 makes "hack" put its result in eax
mov [hack+4], al // xmm{al}
and edx, 15
mov [hack+5], dl // byte [dl] of xmm reg
call hack
pinsrb xmm?, eax, ? // put value back somewhere
...
hack:
db 66 0F 3A 14 00 00 // pextrb ?, ? ,?
ret
As far as I know, you can't do that with full ymm registers (yet?). With some more effort, you could extend it to xmm8-xmm15. It's easily adjustable to other "pointer" sizes and other element sizes.

Two's Complement -- How are negative numbers handled?

It is my understanding that numbers are negated using the two's compliment, which to my understanding is: !num + 1.
So my question is does this mean that, for variable 'foo'=1, a negated 'foo' will be the exactly the same as variable 'bar'=255.
f we were to check if -'foo' == 'bar' or if -'foo' == 255, would we get that they are equal?
I know that some languages, such as Java, keep a sign bit -- so the comparisons would yield false. What of languages that do not? And I'm assuming that assembler/native machine does not have a sign bit.
In addition to all of this, I read about a zero flag or a carry-over flag that is set when a 'negative' number is added to another (of any sign) number. This flag being set whenever it is added because of the way two's complement works, 0x01 + 0xff = 0x00 (with the leading 1 truncated). What exactly is this flag used for?
And my last question, for other math operations (such as multiplication), would I have to re-negate the number (so it is now positive), perform the operation, and negate the result? E.g., !((!neg + 1) * pos) + 1.
Edit
Finished the question, so feel free fire away.
Yes, in two’s complement, the number x is represented as ~x+1, where ~x is the bitwise complement of the binary numeral for x in some fixed number bits. E.g., for eight bits, the binary numeral for x is 000000001, so the bitwise complement is 11111110, and adding one produces 11111111.
There is no way to distinguish -1 in eight-bit two’s complement from 255 in eight-bit binary (with no sign). They both have the same representation in bits: 11111111. If you are using both of these numbers, you must either separately remember which one is eight-bit two’s complement and which one is plain eight-bit binary or you must use more than eight bits. In other words, at the raw bit level, 11111111 is just eight bits; it has no value until we decide how to interpret it.
Java and typical other languages do not maintain a sign bit separate from the value of a number; the sign is part of the encoding of the number. Also, typical languages do not allow you to compare different types. If you have a two’s complement x and an unsigned y, then either one must be converted to the type of the other before comparison or they must both be converted to a third type. Thus, if you compare x and y, and one is converted to the other, then the conversion will overflow or wrap, and you cannot expect to get the correct mathematical result. To compare these two numbers, we might convert each of them to a wider integer, such as 32-bits, then compare. Converting the eight-bit two’s complement 11111111 to a 32-bit integer produces -1, and converting the eight-bit plain binary 11111111 to a 32-bit integer produces 255, and then the comparison reports they are unequal.
The zero flag and the carry flag you read about are flags that are set when a comparison instruction is executed in a computer processor. Most high-level languages do not give you direct access to these flags. Many processors have an instruction with a form like this:
cmp a, b
That instruction subtracts b from a and discards the difference but remembers several flags that describe the subtraction: Was the result zero (zero flag)? Did a borrow occur (borrow flag)? Was the result negative (sign flag)? Did an overflow occur (overflow flag)?
The compare instruction requires that the two things being compared be the same type (two’s complement or unsigned), but it does not care which type. The results can be tested later by checking particular combinations of the flags depending on the type. That is, the information recorded in the flags can distinguish whether one two’s complement number was greater than another or whether one unsigned number was greater than another, depending on what tests are made. There are conditional branch instructions that test the desired flag properties.
There is generally no need to “un-negate” a number to perform arithmetic operations. Processors include arithmetic instructions that work on two’s complement numbers. Usually the add and subtract instructions are type-agnostic, the same way the compare instruction is, but the multiply and divide instructions are not (except for some forms of multiply that return partial results). The add and subtract instructions can be type-agnostic because the wrapping that occurs in the arithmetic works for both two’s complement and unsigned. However, that wrapping does not work for multiplication and division.

Assembler memory address representation

I'm trying to get into assembler and I often come across numbers in the following form:
org 7c00h
; initialize the stack:
mov ax, 07c0h
mov ss, ax
mov sp, 03feh ; top of the stack.
7c00h, 07c0h, 03feh - What is the name of this number notation? What do they mean? Why are they used over "normal" decimal numbers?
It's hexadecimal, the numeral system with 16 digits 0-9 and A-F. Memory addresses are given in hex, because it's shorter, easier to read, and the numbers that represent memory locations don't mean anything special to humans, so no sense to have long numbers. I would guess that somewhere in the past someone had to type in some addresses by hand as well, might as well have started there.
Worth noting also, 0:7C00 is the boot sector load address.
Further worth noting: 07C0:03FE is the same address as 0:7FFE due to the way segmented addressing works.
This guy's left himself a 510 byte stack (he made the very typical off-by-two error in setting up the boot sector's stack).
These are numbers in hexadecimal notation, i.e. in base 16, where A to F have the digit values 10 to 15.
One advantage is that there is a more direct conversion to binary numbers. With a little bit of practice it is easy to see which bits in the number are 1 and which are 0.
Another is is that many numbers used internally, such as memory addresses, are round numbers in hexadecimal, i.e. contain a lot of zeros.

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