Are the usize/isize types in Rust guaranteed to always be either 32 or 64 bits?

Question

If I compile a program targeting a 16 bit architecture, will std::mem::size_of::<usize>() return 2, or is it guaranteed (via a specification, RFC, or otherwise) to always return 4 or 8?

Answer 1

This is what The Rust Reference has to say about usize:

usize and isize have a size big enough to contain every address on the target platform. For example, on a 32 bit target, this is 4 bytes and on a 64 bit target, this is 8 bytes.

Note that the phrasing doesn't exclude sizes other than 4 bytes or 8 bytes. Indeed, Rust already supports a platform with 16-bit usize: msp430-none-elf (the MSP430 is a 16-bit microcontroller).

If you want to perform conditional compilation based on the size of pointers, you can use the target_pointer_width configuration option. Here's a sample usage from the core library:

#[cfg(target_pointer_width = "16")]
#[lang = "usize"]
impl usize {
    uint_impl! { usize, u16, 16, 65535, "", "", 4, "0xa003", "0x3a", "0x1234", "0x3412", "0x2c48",
        "[0x34, 0x12]", "[0x12, 0x34]",
        usize_isize_to_xe_bytes_doc!(), usize_isize_from_xe_bytes_doc!() }
}
#[cfg(target_pointer_width = "32")]
#[lang = "usize"]
impl usize {
    uint_impl! { usize, u32, 32, 4294967295, "", "", 8, "0x10000b3", "0xb301", "0x12345678",
        "0x78563412", "0x1e6a2c48", "[0x78, 0x56, 0x34, 0x12]", "[0x12, 0x34, 0x56, 0x78]",
        usize_isize_to_xe_bytes_doc!(), usize_isize_from_xe_bytes_doc!() }
}

#[cfg(target_pointer_width = "64")]
#[lang = "usize"]
impl usize {
    uint_impl! { usize, u64, 64, 18446744073709551615, "", "", 12, "0xaa00000000006e1", "0x6e10aa",
        "0x1234567890123456", "0x5634129078563412", "0x6a2c48091e6a2c48",
        "[0x56, 0x34, 0x12, 0x90, 0x78, 0x56, 0x34, 0x12]",
         "[0x12, 0x34, 0x56, 0x78, 0x90, 0x12, 0x34, 0x56]",
        usize_isize_to_xe_bytes_doc!(), usize_isize_from_xe_bytes_doc!() }
}