A value is live from where it is written to where it is last read. Two values can share a register only if they are never live at the same time.

When too many values are live at one point, some go to memory: they are spilled. A spill costs a store and a load. The best assignment is NP-hard to find2, so all practical allocators use heuristics.

The calling convention adds two rules. A call can overwrite the caller-saved registers (rax rcx rdx rsi rdi r8-r11 and all xmm registers on Linux). A function must restore the callee-saved registers (rbx rbp r12-r15) before it returns. As an example, this function keeps y live across a call:

long g(long);
long h(long x, long y) {
    long t = g(x);
    return t + y;
}

Before allocation, rdi, rsi and rax are fixed by the calling convention, and v1-v4 are vregs:

0  v1 = copy rdi      ; x
1  v2 = copy rsi      ; y
2  rdi = copy v1      ; argument of g
3  call g             ; clobbers caller-saved
4  v3 = copy rax      ; t
5  v4 = copy v3
6  v4 = add v4, v2
7  rax = copy v4
8  ret

x86 add writes over its first operand (two-address), so instruction 5 copies t first. After allocation, at -O1:

push rbp
mov  rbp, rsp
sub  rsp, 0x8
push rbx              ; rbx is callee-saved: save it
mov  rbx, rsi         ; y
call g                ; x is already in rdi
add  rax, rbx         ; t stays in rax
pop  rbx
leave
ret

Five of the six copies are gone, and y went to a callee-saved register. No code in the allocator says "put values that cross a call in callee-saved registers". It falls out of the design, and that is my favourite part.

Five steps

The allocator runs five steps per function:

1. Live ranges: number the instructions and find where each vreg is live.

2. Fixed registers: mark where the code uses physical registers

   directly.

3. Coalescing: join vregs that a copy connects into one group (a

   bundle3), so the copy can go away.

4. Picking registers: give each bundle a register, most important

   first.

5. Spilling: give stack slots to bundles with no register, then

   rewrite the code.

Each bundle keeps its register or stack slot for its full lifetime. The allocator never:

• takes a register back from a bundle (no eviction)
• splits a range between a register and memory
• runs a step two times

These parts make real allocators big. My measurements say rat does not miss them much.

Live ranges

Slots

Instruction i gets two slots: it reads its operands at 2i and writes its results at 2i+1. A live range is a sorted list of [start, end] slot segments.

Where a source ends depends on the instruction:

• Copies: the source ends at the read slot, and the destination starts at the write

  slot. In instruction 2, rdi = copy v1, v1 ends at slot 4 and rdi starts at slot 5. They do not overlap, so they can share a register and the copy becomes a no-op.

• Other instructions: a source stays live through the write slot, so a result never overwrites a

  different operand. v2 is written by instruction 1 and last read by the add at instruction 6, so it lives in [3, 13].

Live-out sets

rat finds the vregs that are live-out of each block: a later block can still read them. Many compilers do this with one bitset per block and a fixed-point loop. rat does one vreg at a time instead:

1. The vreg is live into each block that reads it before it writes it.

2. From each such block, a worklist goes back through the predecessors and marks the vreg live-out in

   each.

3. The walk stops at a block that defines the vreg.

The cost grows with the blocks where each vreg is live, not with blocks * vregs.4

Segments and weights

Then rat walks each block backward from its live-out set and makes the segments. The same walk sums a weight per vreg: the cost of its spill.

Each def and each use adds 3d, where d is the loop depth (up to 11):

Holes

A live range can have holes, gaps where the vreg is dead. Blocks are numbered in code order, so a range that skips a block has a hole there:

long f(long* a, long n) {
    for(long i = 0; i < n; ++i)
        if(a[i] < 0)
            a[i] = 0;
    return n * 3;
}

The exit block sits between the loop blocks:

mov  eax, 0x0         ; offset 8*i, rax in the loop
cmp  rdx, rdi
jl   loop
exit:
lea  rax, [rdi+rdi*2] ; n*3 in the hole of rax
ret
loop:
mov  rcx, r8
add  rcx, rax
...
add  rax, 0x8
cmp  rdx, rdi
jl   loop
jmp  exit

The offset in rax is dead in the exit block, so n*3 (one lea) can use rax, which is also the return register. A free win from block order.

Fixed registers

rat numbers its registers 1 to 40, so one U64 holds a set of them. Each slot gets one mask, busy[slot]. A set bit means that register is busy at that slot.

The same backward walk marks the physical registers the code uses directly:

The masks and ranges of h:

instr   0  1  2  3  4  5  6  7  8
slot    rw rw rw rw rw rw rw rw rw
rdi     #. .. .#### .. .. .. .. ..
rsi     ####. .. ## .. .. .. .. ..
rax     .. .. .. ####. .. .. .####
others  .. .. .. ## .. .. .. .. ..
v1 x    .======. .. .. .. .. .. ..
v2 y    .. .================ .. ..
v3+v4 t .. .. .. .. .=========. ..

r and w are the read and write slots. # is busy, = is a live range and . is free. A bar continues across the gap between instructions. "others" is every other caller-saved register.

When a bundle gets a register, rat sets that register's bit in every slot of its live range. After that, vregs and fixed registers are bits in the same masks. Each group of 64 slots also has a summary mask, the OR of its 64 masks, so a long range can skip 64 slots at a time.

Coalescing

A copy between two vregs of the same class is a candidate for coalescing. These copies come from:

• two-address instructions

• phi nodes: a

  value that comes from different blocks at a join point

If the two live ranges do not overlap, the vregs become one bundle. It has the merged segments and the summed weight. rat deletes a copy inside one bundle. In h, v3 is [9, 10] and v4 is [11, 14], so they merge.

• rat sorts the copies by loop depth, deepest first. Hot copies merge before cold copies can block them.
• The bundles are kept in a union-find.

• A merge first walks both segment lists to check for overlap. rat skips a merge when the two bundles

  together have more than 256 segments.

A copy between a vreg and a physical register sets a hint instead: the bundle prefers that register if it is free.

Picking registers

Each bundle gets a priority:

priority = weight / sqrt(length in slots)

• Short, hot ranges come first: they matter most and are the easiest to place.
• Long, cold ranges come last and get spilled.
• sqrt keeps a long loop counter from losing too much priority.

rat calls pick on each bundle in priority order:

// cls: register class, gp or xmm
PhysReg pick(VReg v) {
    U64 blocked = ~allocatable[cls];
    for(auto [start, end] : segs[v])
        for(I32 s = start; s <= end; ++s)
            blocked |= busy[s]; // or 64 at a time
    if(hint[v] != kNoReg && !(blocked >> hint[v] & 1))
        return hint[v];
    // caller-saved first, callee-saved last
    return firstFree(order[cls], blocked);
}

Picking in h

In the diagram, rdi is busy only before and after v1, so v1 gets it. Both copies become mov rdi, rdi, and the peephole pass deletes them after allocation.

v2 crosses the call. Every caller-saved register is busy in the call slots, so the first free register is rbx, the first callee-saved one. The prologue saves only the callee-saved registers rat used.

On Linux, no xmm register is callee-saved, so a float that crosses a call always goes to the stack.

Spilling

A bundle with no free register is spilled for its full lifetime. Then:

1. rat sorts the spilled bundles by start. It reuses a stack slot when the last bundle in it has ended.
2. rat rewrites the code. A vreg whose bundle got a register becomes that register.

3. Before each instruction, rat loads each spilled operand into a temporary register. After it, rat stores

   each spilled result.

The temporary is r10 or r11 (xmm14 or xmm15 for floats). No bundle ever gets these. If both are busy, rat takes the first register free at that instruction.

Two cases need no temporary. A copy between a register and a spilled bundle becomes the load or the store itself. A call reads a spilled stack argument from its stack slot directly.

In p, 14 values are live at once:

void p(long* a) {
    long x0 = a[0], x1 = a[1], ..., x13 = a[13];
    a[0] = x0 * x13; a[1] = x1 * x12; a[2] = x2 * x11;
    a[3] = x3 * x10; a[4] = x4 * x9;  a[5] = x5 * x8;
    a[6] = x6 * x7;
}

16 registers minus rsp, rbp, r10, r11 and rdi (which holds a) leaves 11 for 14 values. x0-x6 also hold the products (two-address imul), so they have more uses. Of x7-x13, the three with the longest ranges go to the stack.

Before the peephole pass:

mov  r12, [rdi+0x30]  ; x6, in a register
mov  r10, [rdi+0x38]  ; x7, spilled
mov  [rbp-0x8], r10
mov  r10, [rdi+0x40]  ; x8, spilled
mov  [rbp-0x10], r10
mov  r10, [rdi+0x48]  ; x9, spilled
mov  [rbp-0x18], r10
...
mov  r10, [rbp-0x18]  ; reload x9
imul r9, r10
mov  r10, [rbp-0x10]  ; reload x8
imul rbx, r10
mov  r10, [rbp-0x8]   ; reload x7
imul r12, r10

No instruction between the store of x9 and its reload writes r10. So the peephole pass deletes the reload. Then nothing reads that stack slot, so it also deletes the store.

It can be dumb

Without eviction, an early decision is final. Here is the case that annoys me most:

long sum(long* a, long n) {
    long s = 0;
    for(long i = 0; i < n; ++i)
        s += a[i];
    return s;
}

rat compiles it to:

mov  r9, rdi          ; a: rdi was taken by a[i]
mov  r8, rsi          ; n: rsi was taken by s
...
exit:
mov  rax, rsi         ; s: rax was taken by a+8*i
ret
loop:
mov  rax, r9
add  rax, rcx         ; rax = a + 8*i
mov  rdi, [rax]       ; rdi = a[i]
add  rsi, rdi
...

The loop values are short and hot, so they go first:

1. The address a+8*i takes rax.
2. s loses its hint rax and takes rsi.
3. a[i] takes rdi.
4. a and n come last and lose their hints too.

The result is three movs, all outside the loop.5 An allocator with eviction would fix this chain. I decided three cold movs are not worth the extra code.

Numbers

Against the old allocator:

What each feature was worth

Before the rewrite, I turned off each old feature in turn and measured the code. This was the most useful hour of the project:

Wrapping up

No eviction, no splitting, no second pass, and the new allocator still beats the old one. Most of the quality comes from cheap things: coalescing, hints, holes and use weights.

The lesson for me: measure the old code before I port it. Much of the old allocator did nothing.

References

• Poletto and Sarkar, Linear scan register allocation: the base of the old allocator.
• Max Bernstein, Linear scan register allocation on SSA and Linear scan with lifetime holes: a readable pair of posts.

• Jakob Stoklund Olesen,

  Greedy register allocation in LLVM 3.0: the big version of this idea, with eviction and splitting.

• Chris Fallin, Cranelift, part 4: a new register allocator: a long, good read on bundles.
• Matt Keeter, The solid-state register allocator: even smaller, it runs in one backward pass.

Notes

1. On Windows, only xmm0-xmm3 can be used. xmm4 and

   xmm5 are the spill temporaries, and rat does not use the callee-saved xmm6-xmm15. [back]

2. Chaitin et al.

   showed that any graph can be the interference graph of some program. So register allocation is at least as hard as graph coloring. [back]

3. Cranelift's regalloc2 uses the same word for the same idea. [back]

4. Each block stores the last vreg that marked it, so the walk never clears a visited

   array. [back]

5. The mov inside the loop, mov rax, r9, has a different

   cause. It is the two-address copy for the add, and it always stays. [back]