Can I print -0.000 as 0.000?

Question

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

This outputs -0.000, but shouldn't it be 0.000?

How can I make this print without the minus sign, i.e. 0.000 ?

Answer 1

The correct answer has already been given. Just wanted to add that you'll get the same from c++ cout

If you want to get rid of the sign, it can be done like this:

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;
}

output:

-0.000
-0.000
0.000

EDIT Could this be done without converting to string?

How about:

#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;
}

output:

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

so it seems to work!

But it won't work for all rounding modes.

Therefore it seems safer to use the first version with string convert.

Answer 2

C++ inherits printf from C. In C11 standard §7.21.6.1 The fprintf function, the footnote says:

The results of all floating conversions of a negative zero, and of negative values that round to zero, include a minus sign.

Answer 3

The float representation ( https://en.wikipedia.org/wiki/Floating_point) has a sign bit. Who implemented the printf decided to explicitly put it there even when all the printed numbers are 0.

Answer 4

EDIT: My previous solution didn't solve the problem as it should.

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");
}

Result: printf: -0.000 round : 0.000

Comparing printf w/ precision and 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");
}

Result:

printf: 12345.68
printf: 12345.12

round : 12345.68
round : 12345.12

floor : 12345.67
floor : 12345.12

ceil  : 12345.68
ceil  : 12345.13

So printf seems to also round values when using precision.

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

Old answer, Please ignore:

You might use this to round your answer before printing it:

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

Alternatively, if you just want to cut of the rest:

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

This works even with a negative precision:

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
}

Alternative solution, that allows defining the method of rounding:

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
}