CODEWITHSUNDEEP
c++floating-pointcout
zennehoy
zennehoy

Reputation: 6846

Output float as three digits, or more to avoid exponent

I'm trying to output a float as three digits, or more if necessary to avoid an exponent.

Some examples:

0.12 // 0.123 would be ok
1.23
12.3
123
1234
12345

The closest I've gotten is

std::cout << std::setprecision(3) << f << std::cout;

but this prints things like

21       // rather than 21.0
1.23e+03 // rather than 1234

Combining std::setprecision with std::fixed means I always get the same number of post-decimal digits, which is not what I want.

Using std::setw, 123.456 would still print as 123.456 rather than 123.

Any suggestions?

Upvotes: 3

Views: 244

Answers (2)

Sam Cristall
Sam Cristall

Reputation: 4387

As far as I can tell, the easiest way around this is to roll a function to catch it. I threw this together and it seems to work. I'm not sure if you wanted large numbers to only have 3 significant digits or if they should keep all sig figs to the left of the decimal place, but it wouldn't be hard to make that modification:

void printDigits(float value, int numDigits = 3)
{
    int log10ofValue = static_cast<int>(std::log10(std::abs(value)));
    if(log10ofValue >= 0) //positive log means >= 1
    {
        ++log10ofValue;  //add 1 because we're culling to the left of the decimal now
        //The difference between numDigits and the log10 will let us transition across the decimal
        // in cases like 12.345 or 1.23456 but cap it at 0 for ones greater than 10 ^ numDigits
        std::cout << std::setprecision(std::max(numDigits - log10ofValue, 0)); 
    }
    else
    {
        //We know log10ofValue is <= 0, so set the precision to numDigits + the abs of that value
        std::cout << std::setprecision(numDigits + std::abs(log10ofValue));
    }
    //This is a floating point truncate -- multiply up into integer space, use floor, then divide back down
    float truncated = std::floor(value * std::pow(10.0, numDigits - log10ofValue)) / std::pow(10.0, numDigits - log10ofValue);
    std::cout << std::fixed << truncated << std::endl;
}

Test:

int main(void)
{
    printDigits(0.0000000012345);
    printDigits(12345);
    printDigits(1.234);
    printDigits(12.345678);
    printDigits(0.00012345);
    printDigits(123456789);
    return 0;
}

Output:

0.00000000123
12300
1.23
12.3
0.000123
123000000

Upvotes: 2

zennehoy
zennehoy

Reputation: 6846

Here's the solution I came up with. Ugly, but I believe it works.

if(f>=100) {
    std::cout << std::fixed << std::setprecision(0) << f << std::endl;
    std::cout.unsetf(std::ios_base::floatfield);
} else {
    std::cout << std::showpoint << std::setprecision(3) << f << std::noshowpoint << std::endl;
}

If someone knows how to simplify this, please let me know!

Upvotes: 0

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