我手写的取模：

// #define MOD 2147483647
inline unsigned MODDD(ull x){
    x = (x&MOD) + (x >> 31);
    if(x > MOD) x = (x&MOD) + (x >> 31);
    if(x > MOD) x -= MOD;
    return x;
}

正常的取模：

inline unsigned MODDD(ull x){
    return x%MOD;
}

本来以为手写的会快一点，但没想到正常版更快

于是去 Compiler Explorer 看了它们的汇编

手写版：

push    rbp
        mov     rbp, rsp
        mov     QWORD PTR [rbp-8], rdi
        mov     rax, QWORD PTR [rbp-8]
        and     eax, 2147483647
        mov     rdx, rax
        mov     rax, QWORD PTR [rbp-8]
        shr     rax, 31
        add     rax, rdx
        mov     QWORD PTR [rbp-8], rax
        mov     eax, 2147483648
        cmp     QWORD PTR [rbp-8], rax
        jb      .L2
        mov     rax, QWORD PTR [rbp-8]
        and     eax, 2147483647
        mov     rdx, rax
        mov     rax, QWORD PTR [rbp-8]
        shr     rax, 31
        add     rax, rdx
        mov     QWORD PTR [rbp-8], rax
.L2:
        mov     eax, 2147483648
        cmp     QWORD PTR [rbp-8], rax
        jb      .L3
        sub     QWORD PTR [rbp-8], 2147483647
.L3:
        mov     rax, QWORD PTR [rbp-8]
        pop     rbp
        ret

正常版：

        push    rbp
        mov     rbp, rsp
        mov     QWORD PTR [rbp-8], rdi
        mov     rcx, QWORD PTR [rbp-8]
        movabs  rdx, 8589934597
        mov     rax, rcx
        mul     rdx
        mov     rax, rcx
        sub     rax, rdx
        shr     rax
        add     rax, rdx
        shr     rax, 30
        mov     rdx, rax
        sal     rdx, 31
        sub     rdx, rax
        mov     rax, rcx
        sub     rax, rdx
        pop     rbp
        ret

那么，编译器优化正常版的原理是什么，以及手写能不能达到这样的效果/yiw

悬关orz