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?
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
you can call malloc(1) ofc, but calling malloc_usable_size(malloc(1)) is giving me 24, so it at least allocated 24 bytes for my 1, plus any tracking overhead
yeah, 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.
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?
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_t is word-sized because at some point it gets put into a word-sized register. That’s dumb.
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?
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
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
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.
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.
Indeed. Padding exists. A bool is still one byte.
…of padding. Jesus. Are you going to claim that
uint16_tis not 2 bytes because it is sometimes followed by padding?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.