Question

我们有一个Web服务，它检索查找信息，然后将数据表返回给用户。我们还想创建数据表的哈希值并将其与数据表一起返回。由于我一次只能从Web服务返回一个项目，这将是我可以返回两者的最简单方法。我不想创建列表或类，而是想要数据表。而且我不确定我是否想在数据表中包含哈希... 现在我有一个调用来检索数据表。然后我再次拨打另一个电话来获取哈希值，然后我会比较它是否正确。我这样做是为了验证数据完整性，只是想知道是否有更好的选择。

Answer 1

您的方法必须返回具有这两个属性的类对象。该服务将创建一个新的double对象并填充属性并返回此类实例。

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdint.h>

struct rational {
    uint64_t n;
    uint64_t d;
    unsigned char sign : 1;
};

double uint64_as_double (uint64_t a)
{
    double res;
#if defined (__cplusplus)
    memcpy (&res, &a, sizeof (res));
#else /* __cplusplus */
    volatile union {
        double f;
        uint64_t i;
    } cvt;
    cvt.i = a;
    res = cvt.f;
#endif /* __cplusplus */
    return res;
}

#define ADDcc(a,b,cy,t0,t1) (t0=(b), t1=(a), t0=t0+t1, cy=t0<t1, t0=t0)
#define ADDC(a,b,cy,t0,t1) (t0=(b)+cy, t1=(a), t0+t1)
#define SUBcc(a,b,cy,t0,t1) (t0=(b), t1=(a), cy=t1<t0, t1-t0)

double rational2double (struct rational a)
{
    uint64_t dividend, divisor, quot, rem, t0, t1, cy, res, expo;
    int sticky, round, odd, sign, i;

    dividend = a.n;
    divisor = a.d;
    sign = a.sign;

    /* handle special cases */
    if ((dividend == 0) && (divisor == 0)) {
        res = 0xFFF8000000000000ULL; /* NaN INDEFINITE */
    } else if (dividend == 0) {            
        res = (uint64_t)sign << 63; /* zero */
    } else if (divisor == 0) {
        res = ((uint64_t)sign << 63) | 0x7ff0000000000000ULL; /* Inf */
    } 
    /* handle normal cases */
    else {
        quot = dividend;
        rem = 0;
        expo = 0;
        /* normalize operands using 128-bit shifts */
        while (rem < divisor) {
            quot = ADDcc (quot, quot, cy, t0, t1);
            rem = ADDC (rem, rem, cy, t0, t1);
            expo--;
        }
        /* integer bit of quotient is known to be 1 */
        rem = rem - divisor;
        quot = quot + 1;
        /* generate 53 more quotient bits */
        for (i = 0; i < 53; i++) {
            quot = ADDcc (quot, quot, cy, t0, t1);
            rem = ADDC (rem, rem, cy, t0, t1);
            rem = SUBcc (rem, divisor, cy, t0, t1);
            if (cy) {
                rem = rem + divisor;
            } else {
                quot = quot + 1;
            }
        }
        /* round to nearest or even */
        sticky = rem != 0;
        round = quot & 1;
        quot = quot >> 1;
        odd = quot & 1;
        if (round && (sticky || odd)) {
            quot++;
        }
        /* compose normalized IEEE-754 double-precision number */
        res = ((uint64_t)sign << 63) + ((expo + 64 + 1023 - 1) << 52) + quot;
    }
    return uint64_as_double (res);
}

从wcf服务返回数据表和哈希

1 个答案: