Question

printf("%.3lf\n", -0.0001);

这会输出-0.000，但不应该是0.000？

如何在没有减号的情况下进行此打印，即0.000？

Answer 1

C ++从C继承printf。在C11标准§7.21.6.1fprintf函数中，脚注说：

负零的所有浮动转换的结果，以及舍入为零的负值的结果，包括一个减号。

Answer 2

浮点表示（https://en.wikipedia.org/wiki/Floating_point）有一个符号位。谁实现了printf决定明确地把它放在那里，即使所有打印的数字都是0.

Answer 3

已经给出了正确的答案。只是想补充说你将从c ++ cout

获得相同的内容

如果你想摆脱这个标志，可以这样做：

double fixSign(double d)
{
    std::ostringstream strs;
    strs << std::fixed << std::setprecision(3) << d;
    std::string str = strs.str();

    if (str == "-0.000") return 0.0;

    return d;
}

int main()
{
    double d=-0.0001;

    printf("%.3lf\n", d);

    cout << std::fixed << std::setprecision(3) << d << endl;

    cout << std::fixed << std::setprecision(3) << fixSign(d) << endl;

    return 0;
}

输出：

-0.000
-0.000
0.000

修改的这可以在不转换为字符串的情况下完成吗？

怎么样：

#define PRE 3
#define LIMIT -0.0005  // Must have PRE zeros after the decimal point

// VERSION WITHOUT USE OF STRING
double fixSign_v2(double d)
{
    if ((d < 0) && (d > LIMIT)) return 0;

    return d;
}

double fixSign(double d)
{
    std::ostringstream strs;
    strs << std::fixed << std::setprecision(PRE) << d;
    std::string str = strs.str();

    if (str == "-0.000") return 0.0;

    return d;
}

int main()
{
    // PRE == 2
    //double d1=-0.005;
    //double d2=-0.0049999999999;

    // PRE == 3
    double d1=-0.0005;
    double d2=-0.000499999999999;

    // PRE == 10
    //double d1=-0.00000000005;
    //double d2=-0.0000000000499999999;

    cout << std::fixed << std::setprecision(PRE+20) << d1 << endl;

    cout << std::fixed << std::setprecision(PRE) << fixSign(d1) << endl;

    cout << std::fixed << std::setprecision(PRE) << fixSign_v2(d1) << endl;

    cout << "------------------------" << endl;

    cout << std::fixed << std::setprecision(PRE) << d2 << endl;

    cout << std::fixed << std::setprecision(PRE) << fixSign(d2) << endl;

    cout << std::fixed << std::setprecision(PRE) << fixSign_v2(d2) << endl;

    return 0;
}

输出：

-0.001
-0.001
-0.001
------------------------
-0.000
0.000
0.000

所以它似乎有效！

但它不适用于所有舍入模式。

因此，使用带字符串转换的第一个版本似乎更安全。

Answer 4

编辑：我以前的解决方案并没有解决问题。

template <int precision=0, double(*round_func)(double)=std::round>
double round(double a)
{
    auto multiplier = std::pow(10, precision);
    a *= multiplier;
    a = round_func(a);
    if (a == 0.0) return 0.0;
    a /= multiplier;
    return a;
}

int main()
{
    printf("printf: %.3lf\n", -0.0001);
    printf("round : %.3lf\n", round<3>(-0.0001));
    system("PAUSE");
}

结果： printf：-0.000 回合：0.000

比较printf w / precision和round

int main()
{
    printf("printf: %05.2lf\n", 12345.6789);                        
    printf("printf: %05.2lf\n", 12345.1234);                        
    printf("\n");
    printf("round : %05.2lf\n", round<2, std::round>(12345.6789));  
    printf("round : %05.2lf\n", round<2, std::round>(12345.1234));
    printf("\n");

    printf("floor : %05.2lf\n", round<2, std::floor>(12345.6789));  
    printf("floor : %05.2lf\n", round<2, std::floor>(12345.1234));
    printf("\n");

    printf("ceil  : %05.2lf\n", round<2, std::ceil>(12345.6789));   
    printf("ceil  : %05.2lf\n", round<2, std::ceil>(12345.1234));
    system("PAUSE");
}

结果：

printf: 12345.68
printf: 12345.12

round : 12345.68
round : 12345.12

floor : 12345.67
floor : 12345.12

ceil  : 12345.68
ceil  : 12345.13

因此，当使用精度时，printf似乎也会舍入值。

---------------------------

旧答案，请忽略：

您可以在打印之前使用它来回答您的答案：

template <int precision>
double round(double a)
{
    auto multiplier = std::pow(10, precision);
    a *= multiplier;
    a = std::round(a);
    a /= multiplier;
    return a;
}

或者，如果你只是想减少其余部分：

template <int precision>
double round_down(double a)
{
    auto multiplier = std::pow(10, precision);
    a *= multiplier;
    a = std::floor(a);
    a /= multiplier;
    return a;
}

即使是负精度也是如此：

int main()
{
    //Round at precision
    printf("%5.4lf\n", round<4>(12345.6789));           //Output = 12345.6789
    printf("%5.4lf\n", round<3>(12345.6789));           //Output = 12345.6790
    printf("%5.4lf\n", round<2>(12345.6789));           //Output = 12345.6800
    printf("%5.4lf\n", round<1>(12345.6789));           //Output = 12345.7000
    printf("%5.4lf\n", round<0>(12345.6789));           //Output = 12346.0000
    printf("%5.4lf\n", round<-1>(12345.6789));          //Output = 12350.0000
    printf("%5.4lf\n", round<-2>(12345.6789));          //Output = 12300.0000
    printf("%5.4lf\n", round<-3>(12345.6789));          //Output = 12000.0000
    printf("%5.4lf\n", round<-4>(12345.6789));          //Output = 10000.0000
    printf("%5.4lf\n", round<-5>(12345.6789));          //Output = 0.0000

    //Cut off/Round down after precision
    printf("%5.4lf\n", round_down<4>(12345.6789));      //Output = 12345.6789
    printf("%5.4lf\n", round_down<3>(12345.6789));      //Output = 12345.6780
    printf("%5.4lf\n", round_down<2>(12345.6789));      //Output = 12345.6700
    printf("%5.4lf\n", round_down<1>(12345.6789));      //Output = 12345.6000
    printf("%5.4lf\n", round_down<0>(12345.6789));      //Output = 12345.0000
    printf("%5.4lf\n", round_down<-1>(12345.6789));     //Output = 12340.0000
    printf("%5.4lf\n", round_down<-2>(12345.6789));     //Output = 12300.0000
    printf("%5.4lf\n", round_down<-3>(12345.6789));     //Output = 12000.0000
    printf("%5.4lf\n", round_down<-4>(12345.6789));     //Output = 10000.0000
    printf("%5.4lf\n", round_down<-5>(12345.6789));     //Output = 0.0000
}

替代解决方案，允许定义舍入方法：

template <int precision=0, double(*round_func)(double)=std::round>
double round(double a)
{
    auto multiplier = std::pow(10, precision);
    a *= multiplier;
    a = round_func(a);
    a /= multiplier;
    return a;
}

int main()
{
    printf("%05.4lf\n", round(12345.6789));                     //Output = 12345.0000
    printf("%05.4lf\n", round<1>(12345.6789));                  //Output = 12345.7000
    printf("%05.4lf\n", round<1, std::round>(12345.6789));      //Output = 12345.7000
    printf("%05.4lf\n", round<1, std::floor>(12345.6789));      //Output = 12345.6000
    printf("%05.4lf\n", round<1, std::ceil>(12345.6789));       //Output = 12345.7000
}

我可以将-0.000打印为0.000吗？

4 个答案:

---------------------------

旧答案，请忽略：