将浮点值四舍五入小数“n”位的最佳方法如下，在O(1)时间内:-

我们必须将值四舍五入3位，即n=3。所以,

float a=47.8732355;
printf("%.3f",a);

这里有两个问题:

舍入转换 类型转换。

四舍五入转换意味着四舍五入±浮动/双到最近的地板/天花板浮动/双。 也许你的问题到此为止了。 但如果希望返回Int/Long类型，则需要执行类型转换，因此“溢出”问题可能会影响您的解决方案。所以，检查一下函数中的错误

long round(double x) {
   assert(x >= LONG_MIN-0.5);
   assert(x <= LONG_MAX+0.5);
   if (x >= 0)
      return (long) (x+0.5);
   return (long) (x-0.5);
}

#define round(x) ((x) < LONG_MIN-0.5 || (x) > LONG_MAX+0.5 ?\
      error() : ((x)>=0?(long)((x)+0.5):(long)((x)-0.5))

来源:http://www.cs.tut.fi/~jkorpela/round.html

小心地板(x+0.5)。下面是在[2^52,2^53]范围内奇数的情况:

-bash-3.2$ cat >test-round.c <<END

#include <math.h>
#include <stdio.h>

int main() {
    double x=5000000000000001.0;
    double y=round(x);
    double z=floor(x+0.5);
    printf("      x     =%f\n",x);
    printf("round(x)    =%f\n",y);
    printf("floor(x+0.5)=%f\n",z);
    return 0;
}
END

-bash-3.2$ gcc test-round.c
-bash-3.2$ ./a.out
      x     =5000000000000001.000000
round(x)    =5000000000000001.000000
floor(x+0.5)=5000000000000002.000000

这里是http://bugs.squeak.org/view.php?id=7134。使用@konik这样的解决方案。

我自己的健壮版本是这样的:

double round(double x)
{
    double truncated,roundedFraction;
    double fraction = modf(x, &truncated);
    modf(2.0*fraction, &roundedFraction);
    return truncated + roundedFraction;
}

这里给出了避免下限(x+0.5)的另一个原因。

它通常实现为下限(值+ 0.5)。

编辑:它可能不叫四舍五入，因为我知道至少有三种四舍五入算法:四舍五入到零，四舍五入到最接近的整数，以及银行家的四舍五入。你要求的是最接近的整数。

它在cmath中从c++ 11开始提供(根据http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3337.pdf)

#include <cmath>
#include <iostream>

int main(int argc, char** argv) {
  std::cout << "round(0.5):\t" << round(0.5) << std::endl;
  std::cout << "round(-0.5):\t" << round(-0.5) << std::endl;
  std::cout << "round(1.4):\t" << round(1.4) << std::endl;
  std::cout << "round(-1.4):\t" << round(-1.4) << std::endl;
  std::cout << "round(1.6):\t" << round(1.6) << std::endl;
  std::cout << "round(-1.6):\t" << round(-1.6) << std::endl;
  return 0;
}

输出:

round(0.5):  1
round(-0.5): -1
round(1.4):  1
round(-1.4): -1
round(1.6):  2
round(-1.6): -2

正如在评论和其他回答中指出的那样，ISO c++标准库直到ISO c++ 11才添加round()，当时该函数是通过引用ISO C99标准数学库而引入的。

For positive operands in [½, ub] round(x) == floor (x + 0.5), where ub is 223 for float when mapped to IEEE-754 (2008) binary32, and 252 for double when it is mapped to IEEE-754 (2008) binary64. The numbers 23 and 52 correspond to the number of stored mantissa bits in these two floating-point formats. For positive operands in [+0, ½) round(x) == 0, and for positive operands in (ub, +∞] round(x) == x. As the function is symmetric about the x-axis, negative arguments x can be handled according to round(-x) == -round(x).

这导致了下面的压缩代码。它在各种平台上编译成合理数量的机器指令。我观察到gpu上最紧凑的代码，其中my_roundf()需要大约12条指令。根据处理器架构和工具链的不同，这种基于浮点的方法可能比在不同答案中引用的newlib基于整数的实现更快或更慢。

我使用Intel编译器版本13对my_roundf()与newlib roundf()实现进行了详尽的测试，同时使用/fp:strict和/fp:fast。我还检查了newlib版本是否与Intel编译器mathimf库中的roundf()匹配。对于双精度round()不可能进行详尽的测试，但是代码在结构上与单精度实现相同。

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

float my_roundf (float x)
{
    const float half = 0.5f;
    const float one = 2 * half;
    const float lbound = half;
    const float ubound = 1L << 23;
    float a, f, r, s, t;
    s = (x < 0) ? (-one) : one;
    a = x * s;
    t = (a < lbound) ? x : s;
    f = (a < lbound) ? 0 : floorf (a + half);
    r = (a > ubound) ? x : (t * f);
    return r;
}

double my_round (double x)
{
    const double half = 0.5;
    const double one = 2 * half;
    const double lbound = half;
    const double ubound = 1ULL << 52;
    double a, f, r, s, t;
    s = (x < 0) ? (-one) : one;
    a = x * s;
    t = (a < lbound) ? x : s;
    f = (a < lbound) ? 0 : floor (a + half);
    r = (a > ubound) ? x : (t * f);
    return r;
}

uint32_t float_as_uint (float a)
{
    uint32_t r;
    memcpy (&r, &a, sizeof(r));
    return r;
}

float uint_as_float (uint32_t a)
{
    float r;
    memcpy (&r, &a, sizeof(r));
    return r;
}

float newlib_roundf (float x)
{
    uint32_t w;
    int exponent_less_127;

    w = float_as_uint(x);
    /* Extract exponent field. */
    exponent_less_127 = (int)((w & 0x7f800000) >> 23) - 127;
    if (exponent_less_127 < 23) {
        if (exponent_less_127 < 0) {
            /* Extract sign bit. */
            w &= 0x80000000;
            if (exponent_less_127 == -1) {
                /* Result is +1.0 or -1.0. */
                w |= ((uint32_t)127 << 23);
            }
        } else {
            uint32_t exponent_mask = 0x007fffff >> exponent_less_127;
            if ((w & exponent_mask) == 0) {
                /* x has an integral value. */
                return x;
            }
            w += 0x00400000 >> exponent_less_127;
            w &= ~exponent_mask;
        }
    } else {
        if (exponent_less_127 == 128) {
            /* x is NaN or infinite so raise FE_INVALID by adding */
            return x + x;
        } else {
            return x;
        }
    }
    x = uint_as_float (w);
    return x;
}

int main (void)
{
    uint32_t argi, resi, refi;
    float arg, res, ref;

    argi = 0;
    do {
        arg = uint_as_float (argi);
        ref = newlib_roundf (arg);
        res = my_roundf (arg);
        resi = float_as_uint (res);
        refi = float_as_uint (ref);
        if (resi != refi) { // check for identical bit pattern
            printf ("!!!! arg=%08x  res=%08x  ref=%08x\n", argi, resi, refi);
            return EXIT_FAILURE;
        }
        argi++;
    } while (argi);
    return EXIT_SUCCESS;
}