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Pl/1 did it right:
Dcl 1 mybools, 3 bool1 bit(1) unaligned, 3 bool2 bit(1) unaligned, … 3 bool8 bit(1) unaligned;
All eight bools are in the same byte.
Depending on the language
And compiler. And hardware architecture. And optimization flags.
As usual, it’s some developer that knows little enough to think the walls they see around enclose the entire world.
I don’t think so. Apart from dynamically typed languages which need to store the type with the value, it’s always 1 byte, and that doesn’t depend on architecture (excluding ancient or exotic architectures) or optimisation flags.
Which language/architecture/flags would not store a bool in 1 byte?
Apart from dynamically typed languages which need to store the type with the value
You know that depending on what your code does, the same C that people are talking upthread doesn’t even need to allocate memory to store a variable, right?
How does that work?
I think he’s talking about if a variable only exists in registers. In which case it is the size of a register. But that’s true of everything that gets put in registers. You wouldn’t say
uint16_tis word-sized because at some point it gets put into a word-sized register. That’s dumb.
things that store it as word size for alignment purposes (most common afaik), things that pack multiple books into one byte (normally only things like bool sequences/structs), etc
things that store it as word size for alignment purposes
Nope. bools only need to be naturally aligned, so 1 byte.
If you do
struct SomeBools { bool a; bool b; bool c; bool d; };its 4 bytes.
sure, but if you have a single bool in a stack frame it’s probably going to be more than a byte. on the heap definitely more than a byte
but if you have a single bool in a stack frame it’s probably going to be more than a byte.
Nope. - if you can’t read RISC-V assembly, look at these lines
sb a5,-17(s0) ... sb a5,-18(s0) ... sb a5,-19(s0) ...That is it storing the bools in single bytes. Also I only used RISC-V because I’m way more familiar with it than x86, but it will do the same thing.
on the heap definitely more than a byte
Nope, you can happily
malloc(1)and store a bool in it, ormalloc(4)and store 4 bools in it. A bool is 1 byte. Consider this a TIL moment.c++ guarantees that calls to malloc are aligned https://en.cppreference.com/w/cpp/memory/c/malloc .
you can call
malloc(1)ofc, but callingmalloc_usable_size(malloc(1))is giving me 24, so it at least allocated 24 bytes for my 1, plus any tracking overheadyeah, as I said, in a stack frame. not surprised a compiler packed them into single bytes in the same frame (but I wouldn’t be that surprised the other way either), but the system v abi guarantees at least 4 byte alignment of a stack frame on entering a fn, so if you stored a single bool it’ll get 3+ extra bytes added on the next fn call.
computers align things. you normally don’t have to think about it. Consider this a TIL moment.
Fucking lol at the downvoters haha that second sentence must have rubbed them the wrong way for being too accurate.
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I swore I read that mysql dbs will store multiple bools in a row as bit maps in one byte. I can’t prove it though
I recall that tsql will group bit columns into bytes for storage, wouldn’t surprise me if other flavours did something similar.
SIMD Might be the term youre looking for (Single Input Multiple Data)
Wait till you realise the size of SSD sectors
Now store the numbers (array):
0 0 0 1 0 1 1 2
think 8 bytes???
It’s far more often stored in a word, so 32-64 bytes, depending on the target architecture. At least in most languages.
No it isn’t. All statically typed languages I know of use a byte. Which languages store it in an entire 32 bits? That would be unnecessarily wasteful.
C, C++, C#, to name the main ones. And quite a lot of languages are compiled similarly to these.
To be clear, there’s a lot of caveats to the statement, and it depends on architecture as well, but at the end of the day, it’s rare for a
byteorboolto be mapped directly to a single byte in memory.Say, for example, you have this function…
public void Foo() { bool someFlag = false; int counter = 0; ... }The
someFlagandcountervariables are getting allocated on the stack, and (depending on architecture) that probably means each one is aligned to a 32-bit or 64-bit word boundary, since many CPUs require that for whole-word load and store instructions, or only support a stack pointer that increments in whole words. If the function were to have multiplebyteorboolvariables allocated, it might be able to pack them together, if the CPU supports single-byte load and store instructions, but the nextintvariable that follows might still need some padding space in front of it, so that it aligns on a word boundary.A very similar concept applies to most struct and object implementations. A single
byteorboolfield within a struct or object will likely result in a whole word being allocated, so that other variables and be word-aligned, or so that the whole object meets some optimal word-aligned size. But if you have multiple less-than-a-word fields, they can be packed together. C# does this, for sure, and has some mechanisms by which you can customize field packing.No, in C and C++ a bool is a byte.
since many CPUs require that for whole-word load and store instructions
All modern architectures (ARM, x86 RISC-V) support byte load/store instructions.
or only support a stack pointer that increments in whole words
IIRC the stack pointer is usually incremented in 16-byte units. That’s irrelevant though. If you store a single bool on the stack it would be 1 byte for the bool and 15 bytes of padding.
A single byte or bool field within a struct or object will likely result in a whole word being allocated, so that other variables and be word-aligned
Again, no. I think you’ve sort of heard about this subject but haven’t really understood it.
The requirement is that fields are naturally aligned (up to the machine word size). So a byte needs to be byte-aligned, 2-bytes needs to be 2-byte aligned, etc.
Padding may be inserted to achieve that but that is padding it doesn’t change the size of the actual bool, and it isn’t part of the bool.
But if you have multiple less-than-a-word fields, they can be packed together.
They will be, if it fits the alignment requirements. Create a struct with 8 bools. It will take up 8 bytes no matter what your packing setting is. They even give an example:
If you specify the default packing size, the size of the structure is 8 bytes. The two bytes occupy the first two bytes of memory, because bytes must align on one-byte boundaries.
They used
bytehere but it’s the same forboolbecause a bool is one byte.I’m really surprised how common this misconception is.
It’s not wasteful, it’s faster. You can’t read one byte, you can only read one word. Every decent compiler will turn booleans into words.
You can’t read one byte
lol what. You can absolutely read one byte: https://godbolt.org/z/TeTch8Yhd
On ARM it’s
ldrb(load register byte), and on RISC-V it’slb(load byte).Every decent compiler will turn booleans into words.
No compiler I know of does this. I think you might be getting confused because they’re loaded into registers which are machine-word sized. But in memory a
boolis always one byte.Sorry, but you’re very confused here.
You said you can’t read one byte. I showed that you can. Where’s the confusion?
Internally it will still read a whole word. Because the CPU cannot read less than a word. And if you read the ARM article you linked, it literally says so.
Thus any compiler worth their salt will align all byte variables to words for faster memory access. Unless you specifically disable such behaviour. So yeah, RTFM :)
Wrong again. It depends on the CPU. They can absolutely read a single byte and they will do if you’re reading from non-idempotent memory.
If you’re reading from idempotent memory they won’t read a byte or a word. They’ll likely read a whole cache line (usually 64 bytes).
And if you read the ARM article you linked, it literally says so.
Where?
Thus any compiler worth their salt will align all byte variables to words for faster memory access.
No they won’t because it isn’t faster. The CPU will read the whole cache line that contains the byte.
RTFM
Well, I would but no manual says that because it’s wrong!
Redundancy is nice in the event of bitflip errors
Is the redundancy used for bools? I mean in actual practice.
C/C++ considers an nonzero number, as your true value but false is only zero. This would allow you to guard against going from true to false via bit flip but not false to true.
Other languages like rust define 0 to be false and 1 to be true and any other bit pattern to be invalid for bools.iunno ¯_(ツ)_/¯
Does anybody ever figure in parity when comparing bit sizes and all that jazz or are we only ever concerned with storage space?
You can’t store data in parity bits… so it’s irrelevant.
Back in the day when it mattered, we did it like
#define BV00 (1 << 0) #define BV01 (1 << 1) #define BV02 (1 << 2) #define BV03 (1 << 3) ...etc #define IS_SET(flag, bit) ((flag) & (bit)) #define SET_BIT(var, bit) ((var) |= (bit)) #define REMOVE_BIT(var, bit) ((var) &= ~(bit)) #define TOGGLE_BIT(var, bit) ((var) ^= (bit)) ....then... #define MY_FIRST_BOOLEAN BV00 SET_BIT(myFlags, MY_FIRST_BOOLEAN)With embedded stuff its still done like that. And if you go from the arduino functionss to writing the registers directly its a hell of a lot faster.
Okay. Gen z programmer here. Can you explain this black magic? I see it all the time in kernel code but I have no idea what it means.
The code is a set of preprocessor macros to stuff loads of booleans into one int (or similar), in this case named ‘myFlags’. The preprocessor is a simple (some argue too simple) step at the start of compilation that modifies the source code on its way to the real compiler by substituting #defines, prepending #include’d files, etc.
If myFlags is equal to, e.g. 67, that’s 01000011, meaning that BV00, BV01, and BV07 are all TRUE and the others are FALSE.
The first part is just for convenience and readability. BV00 represents the 0th bit, BV01 is the first etc. (1 << 3) means 00000001, bit shifted left three times so it becomes 00001000 (aka 8).
The middle chunk defines macros to make bit operations more human-readable.
SET_BIT(myFlags, MY_FIRST_BOOLEAN)gets turned into((myFlags) |= ((1 << 0))), which could be simplified asmyFlags = myFlags | 00000001. (Ignore the flood of parentheses, they’re there for safety due to the loaded shotgun nature of the preprocessor.)It’s called bitshifting and is used to select which bits you want to modify so you can toggle them individually.
1 << 0 is the flag for the first bit
1 << 1 for the second
1 << 2 for the third and so onI think that’s correct. It’s been years since I’ve used this technique tbh 😅
Which part?
Edit - oops, responded to wrong comment…
True.
Well storing that would only take half a bit.
Are you telling me that no compiler optimizes this? Why?
Well there are containers that store booleans in single bits (e.g.
std::vector<bool>- which was famously a big mistake).But in the general case you don’t want that because it would be slower.
Why is this a big mistake? I’m not a c++ person
The mistake was that they created a type that behaves like an array in every case except for
bool, for which they created a special magical version that behaves just subtly different enough that it can break things in confusing ways.Could you provide an example?
The biggest problem is that each element doesn’t have a unique memory address; iterators aren’t just pointers.
Consider what the disassembly would look like. There’s no fast way to do it.
It’s also unnecessary since 8 bytes is a negligible amount in most cases. Serialization is the only real scenario where it matters. (Edit: and embedded)
In embedded, if you are to the point that you need to optimize the bools to reduce the footprint, you fucked up sizing your mcu.
CPUs don’t read one bit a a time.
It would be slower to read the value if you had to also do bitwise operations to get the value.
But you can also define your own bitfield types to store booleans packed together if you really need to. I would much rather that than have the compiler do it automatically for me.
They do, that’s the optimisation.
I set all 8 bits to 1 because I want it to be really true.
I was programming in assembly for ARM (some cortex chip) and I kid you not the C program we were integrating with required 255, with just 1 it read it as false
01111111 = true
11111111 = negative true = false
Why do alternative facts always gotta show up uninvited to the party? 🥳
So all this time true was actually false and false was actually true ?
Depends on if you are on a big endian or little endian architecture.
Come on man, I’m not gonna talk about my endian publicly
What if it’s an unsigned boolean?
Cthulhu shows up.
Common misconception… Unsigned booleans (ubool) are always 16-bits.
Super true.
Could also store our bools as floats.
00111111100000000000000000000000is true and10111111100000000000000000000000is negative true.Has the fun twist that true & false is true and true | false is false .
negative true = negative non-zero = non-zero = true.
00001111 = maybe
Schrödingers Boolean
10101010 = I don’t know
100001111 = maybe not
00000001 00000000 00001111 10101010
0011 1111 = could you repeat the question
Is this quantum computing? 😜
TIL, 255 is the new 1.
Aka -1 >> 1 : TRUE
But only if you really mean it. If not, it’s a syntax error and the compiler will know.
You jest, but on some older computers, all ones was the official truth value. Other values may also have been true in certain contexts, but that was the guaranteed one.
I have a solution with a bit fields. Now your bool is 1 byte :
struct Flags { bool flag0 : 1; bool flag1 : 1; bool flag2 : 1; bool flag3 : 1; bool flag4 : 1; bool flag5 : 1; bool flag6 : 1; bool flag7 : 1; };Or for example:
struct Flags { bool flag0 : 1; bool flag1 : 1: int x_cord : 3; int y_cord : 3; };I watched a YouTube video where a dev was optimizing unity code to match the size of data that is sent to the cpu using structs just like this.
just like electronic components, they sell the gates by the chip with multiple gates in them because it’s cheaper
That’s a good analogy.
I mean is it really a waste? What’s minimum amount of bits most CPUs read in one cycle.
In terms of memory usage it’s a waste. But in terms of performance you’re absolutely correct. It’s generally far more efficient to check is a word is 0 than to check if a single bit is zero.
Usually the most effective way is to read and write the same amount of bits as the architecture of the CPU, so for 64 bit CPUs it’s 64 bits at once.
