浮子的平均值是否满足机器精度平分的条件？

这是中点方法的实现，它实际上不使用平均值，而是在 A 和 B 之间的中间位置拆分浮点数集。这是基于 https://www.juliabloggers.com/bisecting-floating-point-numbers/ 的，我相信它满足了要求。代码审查是值得赞赏的。

#include <cstdint> 
#include <math.h> 
#include <cstring>
#include <cassert>


double bisect_float(double L, double R) {
        
    // Use the usual float rules for combining non-finite numbers
    if(!isfinite(L) || !isfinite(R)){
        return(L+R);
    }
    
    if (L==R){return(L);}
    

    // Always return 0.0 when inputs have opposite sign and neither is zero
    if ( (L>0.0 && R<0.0) || (L<0.0 && R>0.0) )   {
        return(0.0);
    }
    
    // check whether we need to change the sign
    bool negate = (L < 0.0 || R < 0.0);
        
    // reinterpret floats as integers
    // this gives their ordering index
    uint64_t Li;
    uint64_t Ri;
    assert(sizeof(uint64_t) == sizeof L);
    assert(sizeof(uint64_t) == sizeof R);
    L = abs(L); // remove sign bit
    R = abs(R); // remove sign bit
    memcpy(&Li, &L, sizeof Li);
    memcpy(&Ri, &R, sizeof Ri);
    
    // find the median float representation
    uint64_t res_i;
    if (Li>Ri){
        std::swap(Li, Ri);
    }
    res_i = Li + ((Ri-Li)>>1);
    //  NOTE: (Li>>1) + (Ri>>1) would be wrong for (3,5), as it would be 1+2=3 and hence not the midpoint
    
    // reinterpret median index as float
    double res;
    memcpy(&res, &res_i, sizeof res);
    
    if (negate){
        return(-res);
    } else {
        return(res);
    }   
}