Greetings everybody. I have seen examples of such operations for so many times that I begin to think that I am getting something wrong with binary arithmetic. Is there any sense to perform the following:
byte value = someAnotherByteValue & 0xFF;
I don't really understand this, because it does not change anything anyway. Thanks for help.
P.S.
I was trying to search for information both elsewhere and here, but unsuccessfully.
EDIT:
Well, off course i assume that someAnotherByteValue is 8 bits long, the problem is that i don't get why so many people ( i mean professionals ) use such things in their code. For example in SharpZlib there is:
buffer_ |= (uint)((window_[windowStart_++] & 0xff |
(window_[windowStart_++] & 0xff) << 8) << bitsInBuffer_);
where window_ is a byte buffer.
The most likely reason is to make the code more self-documenting. In your particular example, it is not the size of someAnotherByteValue that matters, but rather the fact that value is a byte. This makes the & redundant in every language I am aware of. But, to give an example of where it would be needed, if this were Java and someAnotherByteValue was a byte, then the line int value = someAnotherByteValue; could give a completely different result than int value = someAnotherByteValue & 0xff. This is because Java's long, int, short, and byte types are signed, and the rules for conversion and sign extension have to be accounted for.
If you always use the idiom value = someAnotherByteValue & 0xFF then, no matter what the types of the variable are, you know that value is receiving the low 8 bits of someAnotherByteValue.
uint s1 = (uint)(initial & 0xffff);
There is a point to this because uint is 32 bits, while 0xffff is 16 bits. The line selects the 16 least significant bits from initial.
Nope.. There is no use in doing this. Should you be using a value that is having its importance more than 8 bits, then the above statement has some meaning. Otherwise, its the same as the input.
If sizeof(someAnotherByteValue) is more than 8 bits and you want to extract the least signficant 8 bits from someAnotherByteValue then it makes sense. Otherwise, there is no use.
No, there is no point so long as you are dealing with a byte. If value was a long then the lower 8 bits would be the lower 8 bits of someAnotherByteValue and the rest would be zero.
In a language like C++ where operators can be overloaded, it's possible but unlikely that the & operator has been overloaded. That would be pretty unusual and bad practice though.
EDIT: Well, off course i assume that
someAnotherByteValue is 8 bits long,
the problem is that i don't get why so
many people ( i mean professionals )
use such things in their code. For
example in Jon Skeet's MiscUtil there
is:
uint s1 = (uint)(initial & 0xffff);
where initial is int.
In this particular case, the author might be trying to convert an int to a uint. The & with 0xffff would ensure that it would still convert Lowest 2 Bytes, even if the system is not one which has a 2 byte int type.
To be picky, there is no guaranty regarding a machine's byte size. There is no reason to assume in a extremely portable program that the architecture byte is 8 bits wide. To the best of my memory, according to the C standard (for example), a char is one byte, short is wider or the same as char, int is wider or the same as short, long is wider or the same as int and so on. Hence, theoretically there can be a compiler where a long is actually one byte wide, and that byte will be, say, 10 bits wide. Now, to ensure your program behaves the same on that machine, you need to use that (seemingly redundant) coding style.
"Byte" # Wikipedia gives examples for such peculiar architectures.
Related
i found those at arduino.h library, and was confused about the lowbyte macro
#define lowByte(w) ((uint8_t) ((w) & 0xff))
#define highByte(w) ((uint8_t) ((w) >> 8))
at lowByte : wouldn't the conversion from WORD to uint8_t just take the low byte anyway? i know they w & 0x00ff to get the low byte but wouldn't the casting just take the low byte ?
at both the low/high : why wouldn't they use little endians, and read with size/offset
i.e. if the w is 0x12345678, high is 0x1234, low is 0x5678, they write it to memory as 78 56 34 12 at say offset x
to read the w, you read to size of word at location x
to read the high, you read byte/uint8_t at location x
to read the low, you read byte/uint8_t at location x + 2
at lowByte : wouldn't the conversion from WORD to uint8_t just take the low byte anyway? i know they w & 0x00ff to get the low byte but wouldn't the casting just take the low byte ?
Yes. Some people like to be extra explicit in their code anyway, but you are right.
at both the low/high : why wouldn't they use little endians, and read with size/offset
I don't know what that means, "use little endians".
But simply aliasing a WORD as a uint8_t and using pointer arithmetic to "move around" the original object generally has undefined behaviour. You can't alias objects like that. I know your teacher probably said you can because it's all just bits in memory, but your teacher was wrong; C and C++ are abstractions over computer code, and have rules of their own.
Bit-shifting is the conventional way to achieve this.
In the case of lowByte, yes the cast to uint8_t is equivalent to (w) & 0xff).
Regarding "using little endians", you don't want to access individual bytes of the value because you don't necessarily know whether your system is using big endian or little endian.
For example:
uint16_t n = 0x1234;
char *p = (char *)&n;
printf("0x%02x 0x%02x", p[0], p[1]);
If you ran this code on a little endian machine it would output:
0x34 0x12
But if you ran it on a big endian machine you would instead get:
0x12 0x34
By using shifts and bitwise operators you operate on the value which must be the same on all implementations instead of the representation of the value which may differ.
So don't operate on individual bytes unless you have a very specific reason to.
I have int64_t value1; int32_t value2;. Then I have *(int32_t*)&value1 = value2;
Could you please describe what *(int32_t*)&value1 mean? Does it mean "the lowest 32 bits of value1" so *(int32_t*)&value1 = value2; means "the first 32 bits of value1 are replaced with those of value2"?
Or am I completely wrong?
That is going to depend on the byte-order of the machine it runs on. It is pretty poor practice.
On a little endian system, which is most Intel hardware, the lowest order byte is first, so the 32 bit SIGNED value will be written to the lower 32 bits of the 64 bit integer. On a big endian system, it would be written to the high order bits.
note that the int is signed. If value2 is a negative value, the resulting 64bit number will not be negative (unless it was already negative).
It also won't change the high order bits of the 64bit int.
I'd say...don't do that?
EDIT
To more directly answer your question, yes, you are right, depending on what you mean by the "first 32 bits". First in whatever order the platform uses, yes.
&value1 -> will give the address of value1
(int32_t*)&value1 -> tells the compiler to treat the address of value1 as a pointer to an int32_t
*(int32_t*)&value1 -> then dereference the pointer, so assigning to this will put the assigned value into the address of value1 as if it were an int32_t.
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.
This is something I have been thinking while reading programming books and in computer science class at school where we learned how to convert decimal values into hexadecimal.
Can someone please tell me what are the advantages of using hexadecimal values and why we use them in programmnig?
Thank you.
In many cases (like e.g. bit masks) you need to use binary, but binary is hard to read because of its length. Since hexadecimal values can be much easier translated to/from binary than decimals, you could look at hex values as kind of shorthand notation for binary values.
It certainly depends on what you're doing.
It comes as an extension of base 2, which you probably are familiar with as essential to computing.
Check this out for a good discussion of
several applications...
https://softwareengineering.stackexchange.com/questions/170440/why-use-other-number-bases-when-programming/
The hexadecimal digit corresponds 1:1 to a given pattern of 4 bits. With experience, you can map them from memory. E.g. 0x8 = 1000, 0xF = 1111, correspondingly, 0x8F = 10001111.
This is a convenient shorthand where the bit patterns do matter, e.g. in bit maps or when working with i/o ports. To visualize the bit pattern for 169d is in comparison more difficult.
A byte consists of 8 binary digits and is the smallest piece of data that computers normally work with. All other variables a computer works with are constructed from bytes. For example; a single character can be stored in a single byte, and a 32bit integer consists of 4 bytes.
As bytes are so fundamental we want a way to write down their value as neatly and efficiently as possible. One option would be to use binary, but then we would need a lot of digits. This takes up a lot of space and can be confusing when many numbers are written in sequence:
200 201 202 == 11001000 11001001 11001010
Using hexadecimal notation, we can write every byte using just two digits:
200 == C8
Also, as 16 is a power of 2, it is easy to convert between hexadecimal and binary representations in your head. This is useful as sometimes we are only interested in a single bit within the byte. As a simple example, if the first digit of a hexadecimal representation is 0 we know that the first four binary digits are 0.
I'm writing an emulator for the 6502, and basically, there are some instructions where there's an offset saved in one of the registers (mostly X and Y) and I'm wondering, since branch instructions use signed 8 bit integers, do the registers keep their values as 8 bit signed? Meaning this:
switch(opcode) {
//Bunch of opcodes
case 0xD5:
//Read the memory area with final address being address + x offset
int rempResult = a - readMemory(address + x);
//Comparing some things, setting/disabling flags
//Incrementing program counter and cycles/ticks
break;
//More opcodes
}
Let's say in this situation that x = 0xEE. In regular binary, this would mean that x = 238. In the 6502 however, the branch instruction uses signed offset for jumping to memory addresses, so I'm wondering, is the 238 interpreted as -18 in this case, or is it just regular unsigned 8 bit value?
It varies.
They're not explicitly signed or unsigned for arithmetic, logical, shift, or load and store operations.
The conditional branches (and the unconditional one on the later 6502 descendants) all take the argument as signed; otherwise loops would be extremely awkward.
zero, x addressing is achieved by performing an 8-bit addition of x to the zero page address, ignoring carry, and reading from the zero page. So e.g.
LDX #-126 ; which is +130 if unsigned
LDA 23, x
Would read from address 23+130 = 153. But had it been 223+130 then the end read would have been from (223 + 130) MOD 256 = 97.
absolute, x/y is unsigned and carry works correctly (but costs an extra cycle)
(zero, x) is much like the direct version in that the offset is signed but the result is always within the zero page. Then the real address is read from there.
(zero), y is unsigned with carry working and costing.
The "sign" is simply the value of the most significant (aka bit 7) in an 8-bit byte.
6502 has support for signed values in these ways:
The N bit in .P - but it really just tells you if the last instruction turned on or off bit 7 of a memory location or register. It was common to use BPL/BMI to do stuff based on bit 7 in a memory location for flag or "boolean" like use.
The V bit of .P which is flipped "when the result of adding two positive numbers overflows and ends up negative, and when the result of adding two negative numbers overflows and ends up positive"
And of course obeying the sign bit for relative branch instructions only, e.g. BEQ with a value with bit 7 set will move to a lower memory location, not a higher one.
Beyond that, whether that bit means anything is completely up to you and your program. What really makes numbers signed or unsigned is how you display the numbers.
The linked article above goes into what one's complement and two's complement is and how it makes the mathematics work without the 6502 having to care too much about the sign.