How to print a C++ double with the correct number of significant decimal digits?

Question

When dealing with floating point values in Java, calling the toString() method gives a printed value that has the correct number of floating point significant figures. However, in C++, printing a float via stringstream will round the value after 5 or less digits. Is there a way to "pretty print" a float in C++ to the (assumed) correct number of significant figures?

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EDIT: I think I am being misunderstood. I want the output to be of dynamic length, not a fixed precision. I am familiar with setprecision. If you look at the java source for Double, it calculates the number of significant digits somehow, and I would really like to understand how it works and/or how feasible it is to replicate this easily in C++.

/*
 * FIRST IMPORTANT CONSTRUCTOR: DOUBLE
 */
public FloatingDecimal( double d )
{
    long    dBits = Double.doubleToLongBits( d );
    long    fractBits;
    int     binExp;
    int     nSignificantBits;

    // discover and delete sign
    if ( (dBits&signMask) != 0 ){
        isNegative = true;
        dBits ^= signMask;
    } else {
        isNegative = false;
    }
    // Begin to unpack
    // Discover obvious special cases of NaN and Infinity.
    binExp = (int)( (dBits&expMask) >> expShift );
    fractBits = dBits&fractMask;
    if ( binExp == (int)(expMask>>expShift) ) {
        isExceptional = true;
        if ( fractBits == 0L ){
            digits =  infinity;
        } else {
            digits = notANumber;
            isNegative = false; // NaN has no sign!
        }
        nDigits = digits.length;
        return;
    }
    isExceptional = false;
    // Finish unpacking
    // Normalize denormalized numbers.
    // Insert assumed high-order bit for normalized numbers.
    // Subtract exponent bias.
    if ( binExp == 0 ){
        if ( fractBits == 0L ){
            // not a denorm, just a 0!
            decExponent = 0;
            digits = zero;
            nDigits = 1;
            return;
        }
        while ( (fractBits&fractHOB) == 0L ){
            fractBits <<= 1;
            binExp -= 1;
        }
        nSignificantBits = expShift + binExp +1; // recall binExp is  - shift count.
        binExp += 1;
    } else {
        fractBits |= fractHOB;
        nSignificantBits = expShift+1;
    }
    binExp -= expBias;
    // call the routine that actually does all the hard work.
    dtoa( binExp, fractBits, nSignificantBits );
}

After this function, it calls dtoa( binExp, fractBits, nSignificantBits ); which handles a bunch of cases - this is from OpenJDK6

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For more clarity, an example: Java:

double test1 = 1.2593;
double test2 = 0.004963;
double test3 = 1.55558742563;
    
System.out.println(test1);
System.out.println(test2);
System.out.println(test3);

Output:

1.2593
0.004963
1.55558742563

C++:

std::cout << test1 << "\n";
std::cout << test2 << "\n";
std::cout << test3 << "\n";

Output:

1.2593
0.004963
1.55559

Answer 1

Is there a way to "pretty print" a float in C++ to the (assumed) correct number of significant figures?

Yes, you can do it with C++20 std::format, for example:

double test1 = 1.2593;
double test2 = 0.004963;
double test3 = 1.55558742563;
std::cout << std::format("{}", test1) << "\n";
std::cout << std::format("{}", test2) << "\n";
std::cout << std::format("{}", test3) << "\n";

prints

1.2593
0.004963
1.55558742563

The default format will give you the shortest decimal representation with a round-trip guarantee like in Java.

Since this is a new feature and may not be supported by some standard libraries yet, you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):

fmt::print("{}", 1.2593);

Disclaimer: I'm the author of {fmt} and C++20 std::format.