let mut variable: Type;
loop {
    variable = value;
}

or

loop {
    let variable: Type = value;
}
  • chickenf622@sh.itjust.works
    20·
    1 day ago

    Side note, don’t fall into the trap of premature optimization. You’ll more than likely end up shooting yourself in the foot for something that usually doesn’t matter in most cases.

  • haroldstork@lemm.eeEnglish
    14·
    1 day ago

    Functionally equivalent, the compiler will optimize lots for you. As the programmer, your focus should be on whether that variable is relevant to the loop’s scope exclusively or not (i.e. if you need to access the variable after the loop has modified it). If not, keep it in the loop so that its easier to read your code.

  • TehPers@beehaw.orgEnglish
    2·
    22 hours ago

    Without context, there’s no reason to compare the performance of these. The compiler is complex enough that what you do in the loop and after the loop matters with regards to optimizations.

    Do you have more context? What’s actually happening in the code?

  • Anh Kagi@jlai.luEnglish
    7·
    1 day ago

    I would say that they are equivalent. If I’m not mistaken, let statements only reserves space on the stack, and this only increments the stack register.

    And on the latter snippet, the compiler would certainly not bother to modify the stack pointer as the type doesn’t change.

    • Giooschi@lemmy.worldEnglish
      1·
      4 hours ago

      let statements only reserves space on the stack

      It is not guaranteed to do that. It could also use a register or be optimized out completly (like in the example you posted in the other comment).

      The stack pointer is also not changed for each local variable, but instead for each function call, so it wouldn’t make a difference anyway.

    • Anh Kagi@jlai.lu
      13·
      15 hours ago

      according to godbolt: https://rust.godbolt.org/z/hP5Y3qMPW

      use rand::random;

pub fn main1() {
    let mut var : u128;
    loop {
        var = random();
    }
}

pub fn main2() {
    loop {
        let var : u128 = random();
    }
}

      compiles to:

      example::main1::h45edf333d7832d08:
.Lfunc_begin8:
        sub     rsp, 24
.LBB8_1:
.Ltmp80:
        mov     rax, qword ptr [rip + rand::random::he3c23ceb967a3e28@GOTPCREL]
        call    rax
        mov     qword ptr [rsp + 8], rdx
        mov     qword ptr [rsp], rax
        jmp     .LBB8_1
.Ltmp81:
.Lfunc_end8:

example::main2::h1a899b25b96d66db:
.Lfunc_begin9:
        sub     rsp, 24
.LBB9_1:
.Ltmp82:
        mov     rax, qword ptr [rip + rand::random::he3c23ceb967a3e28@GOTPCREL]
        call    rax
        mov     qword ptr [rsp + 8], rdx
        mov     qword ptr [rsp], rax
        jmp     .LBB9_1
.Ltmp83:
.Lfunc_end9:
        jmp     .LBB9_1

      So yeah, exactly the same thing.

  • asudox@lemmy.asudox.dev
    315·
    1 day ago

    The second one is more expensive as you constantly are creating a new variable. In the first one you just constantly change the value of a variable.

    • 1rre@discuss.tchncs.de
      11·
      1 day ago

      This is Rust not Python

      They’re both optimised out by the compiler. If you disable compiler optimisations, they’re identical in machine code anyway, unless you introduce a second loop, in which case the first will be more memory efficient as the memory used in the first loop can be reused in the second loop, whereas if you declare the variable outside the loop it can’t (again, without compiler optimisations, which make the whole comparison pointless).