我正在寻找最好的 C 或 C++ 代码来编码和解码来自/至 double/char 的十进制纬度和经度值。我更喜欢将代码从 double 转换为 char[]，反之亦然 而不是 C++ 字符串。

如果您有代码片段，那也很棒。

澄清一下:我需要将字符串 Degrees/Minutes/Seconds 转换为 double 然后再转换回字符串。我有 3 亿条记录，所以速度是一个大问题。

参见:http://en.wikipedia.org/wiki/Geographic_coordinate_conversion

最佳答案

通过电子邮件与 OP(amanda) 合作，我们开发了一个基于大型 switch-case 语句的快速函数。

amanda 报告称其运行速度比他们之前使用的代码快 15 倍。
考虑到这是运行超过 3 亿条记录，这应该是相当可观的时间节省。

我发现这个问题很有趣。

代码如下:

/* WARNING:  These values are very important, as used under the "default" case. */
#define INT_PART 3
#define DEC_PART 2

double Str2LatLong(char* coord)
//double LLStr::Str2LL(char* coord)
{
    int sign = +1;
    double val;

    int i = 0;  /* an index into coord, the text-input string, indicating the character currently being parsed */

    int p[9] = {0,0,1,  /* degrees */
                0,0,1,  /* minutes */
                0,0,1   /* seconds */
               };
    int* ptr = p;   /* p starts at Degrees. 
                       It will advance to the Decimal part when a decimal-point is encountered,
                       and advance to minutes & seconds when a separator is encountered */
    int  flag = INT_PART; /* Flips back and forth from INT_PART and DEC_PART */

    while(1)
    {
        switch (coord[i])
        {
            /* Any digit contributes to either degrees,minutes, or seconds */
            case '0':
            case '1':
            case '2':
            case '3':
            case '4':
            case '5':
            case '6':
            case '7':
            case '8':
            case '9':
                *ptr = 10* (*ptr) + (coord[i] - '0');
                if (flag == DEC_PART)  /* it'd be nice if I could find a clever way to avoid this test */
                {
                    ptr[1] *= 10;
                }
                break;

            case '.':     /* A decimal point implies ptr is on an integer-part; advance to decimal part */
                flag = DEC_PART; /* after encountering a decimal point, we are now processing the Decimal Part */
                ptr++;  /* ptr[0] is now the Decimal piece; ptr[1] is the Denominator piece (powers of 10) */
                break;

            /* A Null terminator triggers return (no break necessary) */
            case '\0':
                val = p[0]*3600 + p[3]*60 + p[6];             /* All Integer math */
                if (p[1]) val += ((double)p[1]/p[2]) * 3600;  /* Floating-point operations only if needed */
                if (p[4]) val += ((double)p[4]/p[5]) *   60;  /* (ditto) */
                if (p[7]) val += ((double)p[7]/p[8]);         /* (ditto) */
                return sign * val / 3600.0;                 /* Only one floating-point division! */

            case 'W':
            case 'S':
                sign = -1;
                break;

            /* Any other symbol is a separator, and moves ptr from degrees to minutes, or minutes to seconds */
            default:
                /* Note, by setting DEC_PART=2 and INT_PART=3, I avoid an if-test. (testing and branching is slow) */
                ptr += flag;
                flag = INT_PART; /* reset to Integer part, we're now starting a new "piece" (degrees, min, or sec). */
        }
        i++;
    }

    return -1.0;  /* Should never reach here! */
}

关于c++ - 什么是最快的 C++ 类或 C 库将纬度和经度从十进制度数转换为字符串并返回，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/4018775/