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

or

loop {
    let variable: Type = value;
}
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[–] 8 points 2 years ago* (last edited 2 years ago) (2 children)

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.

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  • [–] 18 points 2 years ago* (last edited 2 years ago) (1 child)

    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.

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  • [–] 3 points 2 years ago* (1 child)

    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.

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