我推荐@Jens Gustedt十六进制解决方案：使用%a。
OP希望“以最大精度（或至少到最高有效小数）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

但让我们深入挖掘。。
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。假设IEEE 754 double-precision binary。所以OneSeventh = 1.0/7.0的结果如下。还显示了前面和后面可表示的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>中有两个宏家族可以帮助我们。
第一组是要打印的十进制字符串中的 * 有效 * 位数，因此当扫描字符串时，我们得到原始浮点数。这里显示了C规范的 minimum 值和 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到浮点中的 * 有效 * 位数，然后打印FP，仍然保留相同的字符串表示。这里显示了C规范的 minimum 值和 sample C11编译器。我相信可以用C99前。

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

第一组宏似乎满足OP的 * 有效 * 数字的目标。但这个宏并不总是可用的。

#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-字符串（C89后可用的第2组宏），如何确定FP-字符串-FP（C89后可用的第1组宏）的数字？总的来说，加3是结果。
现在，要打印的 * 有效 * 位数是已知的，并通过<float.h>驱动。
为了打印N * 有效 * 十进制数字，可以使用各种格式。
对于"%e"，precision 字段是前导位和小数点后的位数。所以- 1是有序的。注意：此-1不在初始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才能看到所有的 * 有效 * 位。

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

注意：很多都使用"%f"。显示小数点后6位数字的; 6是显示的默认值，而不是数字的精度。

无损打印浮点数的简短答案（这样它们就可以被读回完全相同的数字，除了NaN和Infinity）：

• 如果你的类型是float：使用printf("%.9g", number)。
• 如果你的类型是double：使用printf("%.17g", number)。

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

如果您只对位（resp十六进制模式）感兴趣，则可以使用%a。这保证您：
如果存在以2为底的精确表示形式，则默认精度足以用于值的精确表示形式，否则，默认精度足以区分double类型的值。
我不得不补充说，这是只有自C99。

不，没有这样的printf宽度说明符来以最大精度打印浮点数。让我解释一下原因。
float和double的最大精度为变量，取决于float或double的实际值。
回想一下，float和double以sign.exponent.mantissa存储。这意味着小数比大数**的小数部分所用的位数要多得多。

例如，float可以很容易地区分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（受尾数位数限制）都用于小数点左边的大部分数字。
%.f修饰符只是说明你想从浮点数中打印多少个十进制值。事实上，* 可用的准确性取决于数字的大小 *，这取决于 * 作为程序员 * 来处理。printf不能/不能为您处理。

只需使用来自<float.h>的宏和可变宽度转换说明符（".*"）：

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

在我对一个答案的评论中，我感叹道，我一直希望有某种方法可以以十进制形式打印浮点值中的所有有效数字，就像问题所问的那样。我终于坐下来写了。它并不十分完美，这是一段演示代码，可以打印附加信息，但在我的测试中基本上是有效的。请让我知道如果你（即）任何人）想要驱动它的整个 Package 器程序的副本以用于测试。

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]);
}

我运行了一个小实验来验证使用DBL_DECIMAL_DIG打印确实可以精确地保留数字的二进制表示。事实证明，对于我尝试的编译器和C库，DBL_DECIMAL_DIG确实是所需的位数，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;
}

我用微软的C编译器19.00.24215.1和gcc版本7.4.0 20170516（Debian 6.3.0-18+ deb 9 u1）运行这个。使用少一个十进制数位可以将比较完全相等的数字的数量减半。（我还验证了所使用的rand()确实产生了大约一百万个不同的数字。）下面是详细的结果。

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

GCC

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

据我所知，有一个很好的算法，允许 * 输出到必要的有效位数，这样当扫描字符串时，原始浮点值在大卫Gay编写的dtoa.c中获得 *，这在Netlib上是可用的here（另请参阅相关的paper）。该代码用于例如在Python、MySQL、Scilab和许多其他语言中。

为了补充Stéphane的答案，最先进的算法被称为“Dragon4”和“Grisu”。
在为自己研究这个问题时，我发现了这篇文章的背景信息，相关论文的链接和实现：
https://www.ryanjuckett.com/printing-floating-point-numbers/