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  <title>A quick and easy way to print a float in base-10. - Alexandr Poltavsky, software developer</title>

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<h2>A quick and easy way to print a float in base-10.</h2>

<p>The implementation of <em>printf</em> for the "<em>%f</em>" format specifier is quite involved.
  I invite you to take a look: <a href="https://gcc.gnu.org/bugzilla/attachment.cgi?id=24137"><em>printf_fp.c</em></a>. That's hard.</p>

<p>But there is a hack, that you could use in cases where you don't need an exact number.</p>

<p>The format of a 32-bit floating point in modern systems is
  <img src="images/fp-format.png" alt="float format" title="Floating point number format" /> where b = 2 ( &copy;Wikipedia ).</p>

<p>You could transform the base-2 exponent to base-10, take the integer part, and multiply the rest of the number with
  fractional part. This way you essentially get a base-10 number. I provide a draft implementation that you could use
  and improve.</p>

<pre><code>#include &lt;cstdio&gt;
#include &lt;cmath&gt;

void print_fp( float number, int digits ) {
    union {
        float d;
        int u;
    };

    d = number;

    //deal with sign
    bool minus = u &gt;&gt; 31;
    u = u &amp; ~(1 &lt;&lt; 31);

    //calculate base-10 exponent
    double exp = (u &gt;&gt; 23) - 127;
    double exp10 = exp / log2( 10.0 );

    //adjust the number
    u = (u &amp; ~(0xFF &lt;&lt; 23)) | (127 &lt;&lt; 23);
    double d2 = double(d) * pow( 10.0, exp10 - int(exp10) );
    if( d2 &gt;= 10.0 ) { d2 = d2 / 10.0; exp10+=1.0; }
    if( d2 &lt; 1.0 ) { d2 = d2 * 10.0; exp10-=1.0; }

    static char digits[] = "0123456789";

    //print sign
    if( minus ) putc( '-', stdout );

    //print integer part
    int i = int( d2 );
    putc( digits[i], stdout );
    putc( '.', stdout );
    d2 = d2 - i;

    //print fractional part
    while( digits-- ) {
        d2 = d2 * 10.0;
        int i = int( d2 );
        putc( digits[i], stdout );
        d2 = d2 - i;
    }

    //print exponent part
    //being a little lazy and using printf
    printf("e%d\n", int(exp10));
}


int main() {

    double d;

    //the two doubles below are special, in that a float can't distinguish between them
    //taken from Bruce Dawson post "Float Precision–From Zero to 100+ Digits"

    d = 8.589973e9;
    printf("%.10f\n", d);
    print_fp( d , 10 );

    d = 8.589974e9;
    printf("%.10f\n", d);
    print_fp( d , 10 );

}


Output:

8589973000.0000000000
8.5899735040e9

8589974000.0000000000
8.5899735040e9
</code></pre>

<p><a href="https://github.com/alexpolt/poetry/blob/master/print-fp.cpp">Github</a> <a href="http://ideone.com/QO1fU5">Ideone</a></p>

<p>Be warned that I haven't done error analysis. All calculations are done in double,
  so I expect this to work flawlessly for floats.</p>

<p>I'd like to thank <strong>Bruce Dawson</strong> for his outstanding series on
  <a href="https://randomascii.wordpress.com/category/floating-point/">floating points</a>.</p>

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