Avoid trailing zeroes in printf()

Question

I keep stumbling on the format specifiers for the printf() family of functions. What I want is to be able to print a double (or float) with a maximum given number of digits

Answer 1

Hit the same issue, double precision is 15 decimal, and float precision is 6 decimal, so I wrote to 2 functions for them separately

#include 
#include 
#include 
#include 

std::string doublecompactstring(double d)
{
    char buf[128] = {0};
    if (isnan(d))
        return "NAN";
    sprintf(buf, "%.15f", d);
    // try to remove the trailing zeros
    size_t ccLen = strlen(buf);
    for(int i=(int)(ccLen -1);i>=0;i--)
    {
        if (buf[i] == '0')
            buf[i] = '\0';
        else
            break;
    }

    return buf;
}

std::string floatcompactstring(float d)
{
    char buf[128] = {0};
    if (isnan(d))
        return "NAN";
    sprintf(buf, "%.6f", d);
    // try to remove the trailing zeros
    size_t ccLen = strlen(buf);
    for(int i=(int)(ccLen -1);i>=0;i--)
    {
        if (buf[i] == '0')
            buf[i] = '\0';
        else
            break;
    }

    return buf;
}

int main(int argc, const char* argv[])
{
    double a = 0.000000000000001;
    float  b = 0.000001f;

    printf("a: %s\n", doublecompactstring(a).c_str());
    printf("b: %s\n", floatcompactstring(b).c_str());
    return 0;
}

output is

a: 0.000000000000001
b: 0.000001