How does Rust's 128-bit integer `i128` work on a 64-bit system?

Rust has 128-bit integers, these are denoted with the data type i128 (and u128 for unsigned ints):

let a: i128 = 170141183460469231731687303715884105727;

How does Rust make these i128 values work on a 64-bit system; e.g. how does it do arithmetic on these?

Since, as far as I know, the value cannot fit in one register of a x86-64 CPU, does the compiler somehow use two registers for one i128 value? Or are they instead using some kind of big integer struct to represent them?

Solution 1:

All Rust's integer types are compiled to LLVM integers. The LLVM abstract machine allows integers of any bit width from 1 to 2^23 - 1.* LLVM instructions typically work on integers of any size.

Obviously, there aren't many 8388607-bit architectures out there, so when the code is compiled to native machine code, LLVM has to decide how to implement it. The semantics of an abstract instruction like add are defined by LLVM itself. Typically, abstract instructions that have a single-instruction equivalent in native code will be compiled to that native instruction, while those that don't will be emulated, possibly with multiple native instructions. mcarton's answer demonstrates how LLVM compiles both native and emulated instructions.

(This doesn't only apply to integers that are larger than the native machine can support, but also to those that are smaller. For example, modern architectures might not support native 8-bit arithmetic, so an add instruction on two i8s may be emulated with a wider instruction, the extra bits discarded.)

Does the compiler somehow use 2 registers for one i128 value? Or are they using some kind of big integer struct to represent them?

At the level of LLVM IR, the answer is neither: i128 fits in a single register, just like every other single-valued type. On the other hand, once translated to machine code, there isn't really a difference between the two, because structs may be decomposed into registers just like integers. When doing arithmetic, though, it's a pretty safe bet that LLVM will just load the whole thing into two registers.

* However, not all LLVM backends are created equal. This answer relates to x86-64. I understand that backend support for sizes larger than 128 and non-powers of two is spotty (which may partly explain why Rust only exposes 8-, 16-, 32-, 64-, and 128-bit integers). According to est31 on Reddit, rustc implements 128 bit integers in software when targeting a backend that doesn't support them natively.

Solution 2:

The compiler will store these in multiple registers and use multiple instructions to do arithmetic on those values if needed. Most ISAs have an add-with-carry instruction like x86's adc which makes it fairly efficient to do extended-precision integer add/sub.

For example, given

fn main() {
    let a = 42u128;
    let b = a + 1337;
}

the compiler generates the following when compiling for x86-64 without optimization:
(comments added by @PeterCordes)

playground::main:
    sub rsp, 56
    mov qword ptr [rsp + 32], 0
    mov qword ptr [rsp + 24], 42         # store 128-bit 0:42 on the stack
                                         # little-endian = low half at lower address

    mov rax, qword ptr [rsp + 24]
    mov rcx, qword ptr [rsp + 32]        # reload it to registers

    add rax, 1337                        # add 1337 to the low half
    adc rcx, 0                           # propagate carry to the high half. 1337u128 >> 64 = 0

    setb    dl                           # save carry-out (setb is an alias for setc)
    mov rsi, rax
    test    dl, 1                        # check carry-out (to detect overflow)
    mov qword ptr [rsp + 16], rax        # store the low half result
    mov qword ptr [rsp + 8], rsi         # store another copy of the low half
    mov qword ptr [rsp], rcx             # store the high half
                             # These are temporary copies of the halves; probably the high half at lower address isn't intentional
    jne .LBB8_2                       # jump if 128-bit add overflowed (to another not-shown block of code after the ret, I think)

    mov rax, qword ptr [rsp + 16]
    mov qword ptr [rsp + 40], rax     # copy low half to RSP+40
    mov rcx, qword ptr [rsp]
    mov qword ptr [rsp + 48], rcx     # copy high half to RSP+48
                  # This is the actual b, in normal little-endian order, forming a u128 at RSP+40
    add rsp, 56
    ret                               # with retval in EAX/RAX = low half result

where you can see that the value 42 is stored in rax and rcx.

(editor's note: x86-64 C calling conventions return 128-bit integers in RDX:RAX. But this main doesn't return a value at all. All the redundant copying is purely from disabling optimization, and that Rust actually checks for overflow in debug mode.)

For comparison, here is the asm for Rust 64-bit integers on x86-64 where no add-with-carry is needed, just a single register or stack-slot for each value.

playground::main:
    sub rsp, 24
    mov qword ptr [rsp + 8], 42           # store
    mov rax, qword ptr [rsp + 8]          # reload
    add rax, 1337                         # add
    setb    cl
    test    cl, 1                         # check for carry-out (overflow)
    mov qword ptr [rsp], rax              # store the result
    jne .LBB8_2                           # branch on non-zero carry-out

    mov rax, qword ptr [rsp]              # reload the result
    mov qword ptr [rsp + 16], rax         # and copy it (to b)
    add rsp, 24
    ret

.LBB8_2:
    call panic function because of integer overflow

The setb / test is still totally redundant: jc (jump if CF=1) would work just fine.

With optimization enabled, the Rust compiler doesn't check for overflow so + works like .wrapping_add().