Printf 宽度说明符以保持浮点值的精度

我推荐@Jens Gustedt 十六进制解决方案：使用 %a。

OP 想要“以最大精度（或至少到最重要的小数点）打印”。

一个简单的例子是打印七分之一，如下所示：

#include <float.h>
int Digs = DECIMAL_DIG;
double OneSeventh = 1.0/7.0;
printf("%.*e\n", Digs, OneSeventh);
// 1.428571428571428492127e-01

但是让我们更深入地挖掘......

从数学上讲，答案是“0.142857 142857 142857 ...”，但我们使用的是有限精度浮点数。让我们假设 IEEE 754 double-precision binary。因此 OneSeventh = 1.0/7.0 产生以下值。还显示了前面和后面的可表示的 double 浮点数。

OneSeventh before = 0.1428571428571428 214571170656199683435261249542236328125
OneSeventh        = 0.1428571428571428 49212692681248881854116916656494140625
OneSeventh after  = 0.1428571428571428 769682682968777953647077083587646484375

打印 double 的精确十进制表示的用途有限。

C 在 <float.h> 中有 2 个宏系列来帮助我们。
第一组是在十进制字符串中打印的有效位数，所以当扫描回字符串时，我们得到原始浮点数。显示了 C 规范的 minimum 值和一个 sample C11 编译器。

FLT_DECIMAL_DIG   6,  9 (float)                           (C11)
DBL_DECIMAL_DIG  10, 17 (double)                          (C11)
LDBL_DECIMAL_DIG 10, 21 (long double)                     (C11)
DECIMAL_DIG      10, 21 (widest supported floating type)  (C99)

第二组是字符串可以被扫描成浮点数，然后打印 FP，仍然保留相同的字符串表示形式。显示了 C 规范的最小值和示例 C11 编译器。我相信在 C99 之前可用。

FLT_DIG   6, 6 (float)
DBL_DIG  10, 15 (double)
LDBL_DIG 10, 18 (long double)

第一组宏似乎符合 OP 的有效数字目标。但该宏并不总是可用。

#ifdef DBL_DECIMAL_DIG
  #define OP_DBL_Digs (DBL_DECIMAL_DIG)
#else  
  #ifdef DECIMAL_DIG
    #define OP_DBL_Digs (DECIMAL_DIG)
  #else  
    #define OP_DBL_Digs (DBL_DIG + 3)
  #endif
#endif

“+ 3”是我之前回答的症结所在。它的中心是如果知道往返转换字符串-FP-string（设置 #2 宏可用 C89），如何确定 FP-string-FP 的数字（设置 #1 宏在 C89 后可用）？一般来说，加 3 是结果。

现在要打印多少 有效 位是已知的，并通过 <float.h> 驱动。

要打印 N 个有效十进制数字，可以使用各种格式。

对于 "%e"，precision 字段是前导数字和小数点 之后的位数。所以 - 1 是有序的。注意：这个 -1 不在最初的 int Digs = DECIMAL_DIG; 中

printf("%.*e\n", OP_DBL_Digs - 1, OneSeventh);
// 1.4285714285714285e-01

对于 "%f"，precision 字段是小数点 之后的位数。对于像 OneSeventh/1000000.0 这样的数字，需要 OP_DBL_Digs + 6 才能看到所有有效位数字。

printf("%.*f\n", OP_DBL_Digs    , OneSeventh);
// 0.14285714285714285
printf("%.*f\n", OP_DBL_Digs + 6, OneSeventh/1000000.0);
// 0.00000014285714285714285

注意：很多都习惯于"%f"。小数点后显示 6 位； 6 是显示默认值，而不是数字的精度。

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无损打印浮点数的简短答案（以便可以将它们读回完全相同的数字，除了 NaN 和 Infinity）：

如果您的类型是浮点数：使用 printf("%.9g", number)。

如果您的类型是 double：使用 printf("%.17g", number)。

不要使用 %f，因为它只指定小数点后的有效数字位数，并且会截断小数。作为参考，可以在定义 FLT_DECIMAL_DIG 和 DBL_DECIMAL_DIG 的 float.h 中找到幻数 9 和 17。

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如果您只对位（resp 十六进制模式）感兴趣，您可以使用 %a 格式。这保证您：

如果存在以 2 为底的精确表示，则默认精度足以准确表示该值，否则该精度足以区分 double 类型的值。

我必须补充一点，这仅在 C99 之后可用。

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不，没有这样的 printf 宽度说明符可以以最高精度打印浮点数。让我解释一下为什么。

float 和 double 的最大精度是变量，取决于 float 或 double 的实际值。

召回 float 和 double 以 sign.exponent.mantissa 格式存储。这意味着用于小数的小数部分比用于大数的位数多得多。

https://i.stack.imgur.com/JktFK.png

例如，float 可以轻松区分 0.0 和 0.1。

float r = 0;
printf( "%.6f\n", r ) ; // 0.000000
r+=0.1 ;
printf( "%.6f\n", r ) ; // 0.100000

但是 float 不知道 1e27 和 1e27 + 0.1 之间的区别。

r = 1e27;
printf( "%.6f\n", r ) ; // 999999988484154753734934528.000000
r+=0.1 ;
printf( "%.6f\n", r ) ; // still 999999988484154753734934528.000000

这是因为所有精度（受尾数位数的限制）都用于小数点左侧的大部分数字。

%.f 修饰符只是说明您希望从浮点数中打印多少个十进制值，就格式化 而言。 可用的准确性取决于数字的大小这一事实取决于作为程序员的您来处理。 printf 不能/不能为您处理。

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只需使用 <float.h> 中的宏和可变宽度转换说明符 (".*")：

float f = 3.14159265358979323846;
printf("%.*f\n", FLT_DIG, f);

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在我对答案的评论之一中，我感叹我一直想要某种方法以十进制形式以浮点值打印所有有效数字，这与问题所要求的方式大致相同。好吧，我终于坐下来写了。它不是很完美，这是打印附加信息的演示代码，但它主要适用于我的测试。如果您（即任何人）想要驱动它进行测试的整个包装程序的副本，请告诉我。

static unsigned int
ilog10(uintmax_t v);

/*
 * Note:  As presented this demo code prints a whole line including information
 * about how the form was arrived with, as well as in certain cases a couple of
 * interesting details about the number, such as the number of decimal places,
 * and possibley the magnitude of the value and the number of significant
 * digits.
 */
void
print_decimal(double d)
{
        size_t sigdig;
        int dplaces;
        double flintmax;

        /*
         * If we really want to see a plain decimal presentation with all of
         * the possible significant digits of precision for a floating point
         * number, then we must calculate the correct number of decimal places
         * to show with "%.*f" as follows.
         *
         * This is in lieu of always using either full on scientific notation
         * with "%e" (where the presentation is always in decimal format so we
         * can directly print the maximum number of significant digits
         * supported by the representation, taking into acount the one digit
         * represented by by the leading digit)
         *
         *        printf("%1.*e", DBL_DECIMAL_DIG - 1, d)
         *
         * or using the built-in human-friendly formatting with "%g" (where a
         * '*' parameter is used as the number of significant digits to print
         * and so we can just print exactly the maximum number supported by the
         * representation)
         *
         *         printf("%.*g", DBL_DECIMAL_DIG, d)
         *
         *
         * N.B.:  If we want the printed result to again survive a round-trip
         * conversion to binary and back, and to be rounded to a human-friendly
         * number, then we can only print DBL_DIG significant digits (instead
         * of the larger DBL_DECIMAL_DIG digits).
         *
         * Note:  "flintmax" here refers to the largest consecutive integer
         * that can be safely stored in a floating point variable without
         * losing precision.
         */
#ifdef PRINT_ROUND_TRIP_SAFE
# ifdef DBL_DIG
        sigdig = DBL_DIG;
# else
        sigdig = ilog10(uipow(FLT_RADIX, DBL_MANT_DIG - 1));
# endif
#else
# ifdef DBL_DECIMAL_DIG
        sigdig = DBL_DECIMAL_DIG;
# else
        sigdig = (size_t) lrint(ceil(DBL_MANT_DIG * log10((double) FLT_RADIX))) + 1;
# endif
#endif
        flintmax = pow((double) FLT_RADIX, (double) DBL_MANT_DIG); /* xxx use uipow() */
        if (d == 0.0) {
                printf("z = %.*s\n", (int) sigdig + 1, "0.000000000000000000000"); /* 21 */
        } else if (fabs(d) >= 0.1 &&
                   fabs(d) <= flintmax) {
                dplaces = (int) (sigdig - (size_t) lrint(ceil(log10(ceil(fabs(d))))));
                if (dplaces < 0) {
                        /* XXX this is likely never less than -1 */
                        /*
                         * XXX the last digit is not significant!!! XXX
                         *
                         * This should also be printed with sprintf() and edited...
                         */
                        printf("R = %.0f [%d too many significant digits!!!, zero decimal places]\n", d, abs(dplaces));
                } else if (dplaces == 0) {
                        /*
                         * The decimal fraction here is not significant and
                         * should always be zero  (XXX I've never seen this)
                         */
                        printf("R = %.0f [zero decimal places]\n", d);
                } else {
                        if (fabs(d) == 1.0) {
                                /*
                                 * This is a special case where the calculation
                                 * is off by one because log10(1.0) is 0, but
                                 * we still have the leading '1' whole digit to
                                 * count as a significant digit.
                                 */
#if 0
                                printf("ceil(1.0) = %f, log10(ceil(1.0)) = %f, ceil(log10(ceil(1.0))) = %f\n",
                                       ceil(fabs(d)), log10(ceil(fabs(d))), ceil(log10(ceil(fabs(d)))));
#endif
                                dplaces--;
                        }
                        /* this is really the "useful" range of %f */
                        printf("r = %.*f [%d decimal places]\n", dplaces, d, dplaces);
                }
        } else {
                if (fabs(d) < 1.0) {
                        int lz;

                        lz = abs((int) lrint(floor(log10(fabs(d)))));
                        /* i.e. add # of leading zeros to the precision */
                        dplaces = (int) sigdig - 1 + lz;
                        printf("f = %.*f [%d decimal places]\n", dplaces, d, dplaces);
                } else {                /* d > flintmax */
                        size_t n;
                        size_t i;
                        char *df;

                        /*
                         * hmmmm...  the easy way to suppress the "invalid",
                         * i.e. non-significant digits is to do a string
                         * replacement of all dgits after the first
                         * DBL_DECIMAL_DIG to convert them to zeros, and to
                         * round the least significant digit.
                         */
                        df = malloc((size_t) 1);
                        n = (size_t) snprintf(df, (size_t) 1, "%.1f", d);
                        n++;                /* for the NUL */
                        df = realloc(df, n);
                        (void) snprintf(df, n, "%.1f", d);
                        if ((n - 2) > sigdig) {
                                /*
                                 * XXX rounding the integer part here is "hard"
                                 * -- we would have to convert the digits up to
                                 * this point back into a binary format and
                                 * round that value appropriately in order to
                                 * do it correctly.
                                 */
                                if (df[sigdig] >= '5' && df[sigdig] <= '9') {
                                        if (df[sigdig - 1] == '9') {
                                                /*
                                                 * xxx fixing this is left as
                                                 * an exercise to the reader!
                                                 */
                                                printf("F = *** failed to round integer part at the least significant digit!!! ***\n");
                                                free(df);
                                                return;
                                        } else {
                                                df[sigdig - 1]++;
                                        }
                                }
                                for (i = sigdig; df[i] != '.'; i++) {
                                        df[i] = '0';
                                }
                        } else {
                                i = n - 1; /* less the NUL */
                                if (isnan(d) || isinf(d)) {
                                        sigdig = 0; /* "nan" or "inf" */
                                }
                        }
                        printf("F = %.*s. [0 decimal places, %lu digits, %lu digits significant]\n",
                               (int) i, df, (unsigned long int) i, (unsigned long int) sigdig);
                        free(df);
                }
        }

        return;
}


static unsigned int
msb(uintmax_t v)
{
        unsigned int mb = 0;

        while (v >>= 1) { /* unroll for more speed...  (see ilog2()) */
                mb++;
        }

        return mb;
}

static unsigned int
ilog10(uintmax_t v)
{
        unsigned int r;
        static unsigned long long int const PowersOf10[] =
                { 1LLU, 10LLU, 100LLU, 1000LLU, 10000LLU, 100000LLU, 1000000LLU,
                  10000000LLU, 100000000LLU, 1000000000LLU, 10000000000LLU,
                  100000000000LLU, 1000000000000LLU, 10000000000000LLU,
                  100000000000000LLU, 1000000000000000LLU, 10000000000000000LLU,
                  100000000000000000LLU, 1000000000000000000LLU,
                  10000000000000000000LLU };

        if (!v) {
                return ~0U;
        }
        /*
         * By the relationship "log10(v) = log2(v) / log2(10)", we need to
         * multiply "log2(v)" by "1 / log2(10)", which is approximately
         * 1233/4096, or (1233, followed by a right shift of 12).
         *
         * Finally, since the result is only an approximation that may be off
         * by one, the exact value is found by subtracting "v < PowersOf10[r]"
         * from the result.
         */
        r = ((msb(v) * 1233) >> 12) + 1;

        return r - (v < PowersOf10[r]);
}

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我进行了一个小实验来验证使用 DBL_DECIMAL_DIG 打印确实确实保留了数字的二进制表示。事实证明，对于我尝试过的编译器和 C 库，DBL_DECIMAL_DIG 确实是所需的位数，并且即使少一位打印也会产生严重的问题。

#include <float.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

union {
    short s[4];
    double d;
} u;

void
test(int digits)
{
    int i, j;
    char buff[40];
    double d2;
    int n, num_equal, bin_equal;

    srand(17);
    n = num_equal = bin_equal = 0;
    for (i = 0; i < 1000000; i++) {
        for (j = 0; j < 4; j++)
            u.s[j] = (rand() << 8) ^ rand();
        if (isnan(u.d))
            continue;
        n++;
        sprintf(buff, "%.*g", digits, u.d);
        sscanf(buff, "%lg", &d2);
        if (u.d == d2)
            num_equal++;
        if (memcmp(&u.d, &d2, sizeof(double)) == 0)
            bin_equal++;
    }
    printf("Tested %d values with %d digits: %d found numericaly equal, %d found binary equal\n", n, digits, num_equal, bin_equal);
}

int
main()
{
    test(DBL_DECIMAL_DIG);
    test(DBL_DECIMAL_DIG - 1);
    return 0;
}

我使用 Microsoft 的 C 编译器 19.00.24215.1 和 gcc 版本 7.4.0 20170516 (Debian 6.3.0-18+deb9u1) 运行它。使用少一位十进制数字会使比较完全相等的数字数量减半。 （我还验证了使用的 rand() 确实产生了大约一百万个不同的数字。）这里是详细的结果。

微软 C

Tested 999507 values with 17 digits: 999507 found numericaly equal, 999507 found binary equal
Tested 999507 values with 16 digits: 545389 found numericaly equal, 545389 found binary equal

海合会

Tested 999485 values with 17 digits: 999485 found numericaly equal, 999485 found binary equal
Tested 999485 values with 16 digits: 545402 found numericaly equal, 545402 found binary equal

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据我所知，有一种广泛传播的算法允许输出到必要数量的有效数字，这样当重新扫描字符串时，在 David 编写的 dtoa.c 中获取原始浮点值 Gay，可在 Netlib 上获得 here（另请参阅相关的 paper）。此代码用于 Python、MySQL、Scilab 等许多其他语言中。

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