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How many CPU ticks does it take to store a single None? [closed]

Solutons:


In short: In this case, since X does not implement Drop, storing None is done in a single instruction. The number of CPU cycles depends heavily on the exact hardware and on caching effects. But assuming a modern x86-64 CPU and assuming the memory is in L1 cache (fair assumption for stack memory), it should only be a couple cycles.

To get meaningful assembly out of your program, I added two black_box calls to simulate using the value of your variable. I also made the value inside the array 13 as 0 is often treated in a special case.

let mut x : Option<X> = Some([13; 256]);
std::hint::black_box(&x);
x = None; // how many CPU ticks here?
std::hint::black_box(x);

Compiling that code in the Compiler Explorer gives the following assembly (commented by me):

.LCPI0_0:
    .zero   16,13    ; 16 bytes with value 13

example::foo:
    ; reserve space on the stack
    sub     rsp, 144      

    ; Copy the xmmword (16 bytes) from the constant location LCPI0_0 to the 
    ; register xmm0.
    movaps  xmm0, xmmword ptr [rip + .LCPI0_0]

    ; Copy the register xmm0 (now containing a bunch of 13s) to all these memory 
    ; locations (relative to the rsp register)
    movups  xmmword ptr [rsp + 121], xmm0
    movups  xmmword ptr [rsp + 105], xmm0
    movups  xmmword ptr [rsp + 89], xmm0
    movups  xmmword ptr [rsp + 73], xmm0
    movups  xmmword ptr [rsp + 57], xmm0
    movups  xmmword ptr [rsp + 41], xmm0
    movups  xmmword ptr [rsp + 25], xmm0
    movups  xmmword ptr [rsp + 9], xmm0
    movups  xmmword ptr [rsp - 7], xmm0
    movups  xmmword ptr [rsp - 23], xmm0
    movups  xmmword ptr [rsp - 39], xmm0
    movups  xmmword ptr [rsp - 55], xmm0
    movups  xmmword ptr [rsp - 71], xmm0
    movups  xmmword ptr [rsp - 87], xmm0
    movups  xmmword ptr [rsp - 103], xmm0
    movups  xmmword ptr [rsp - 119], xmm0

    ; Move 1 to that single memory location
    mov     byte ptr [rsp - 120], 1

    ; Prepare parameters for black_box call -> ignore that
    lea     rax, [rsp - 120]
    mov     qword ptr [rsp - 128], rax
    lea     rcx, [rsp - 128]

    ; Write None to the variable  <--- LOOK HERE
    mov     byte ptr [rsp - 120], 0

    ; Prepare parameters for black_box call again -> ignore that
    mov     qword ptr [rsp - 128], rax

    ; Function outro: "clean" stack space and return from function
    add     rsp, 144
    ret

As you can see (assuming you trust my comments), storing a None here takes only one instruction, a mov byte one. This is one of the easiest instructions, but also one of the hardest to estimate the time of. Why? CPU-caching. Assuming the data is in L1 cache, this should be only a couple of CPU cycles. If the memory is not in any cache (unlikely for the stack), then it could take hundreds of cycles.

Also worth noting: this is so simple because the replaced data were just simple bytes and no cleanup was needed. If the T in Option<T> implements Drop, the drop function needs to be called at this point, making it a lot more complicated.

Why can this be done with only one byte stored you ask? Well, enums in Rust (which Option is) are generally stored like this:

┌─────┬─────────────┐  
│ tag │ contents... │  
└─────┴─────────────┘

The tag, or discriminator, just stores what variant of the enum (None or Some) is currently active. After that, the data area begins, where the data belonging to the active variant is stored.

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