Question

If you print a float with more precision than is stored in memory, aren’t the extra places supposed to have zeros in them? I have code that is something like this:

double z[2*N]="0";
...
for( n=1; n<=2*N; n++) {
  fprintf( u1, "%.25g", z[n-1]);
  fputc( n<2*N ? ',' : '\n', u1);
}

Which is creating output like this:

0,0.7071067811865474617150085,....

A float should have only 17 decimal places (right? Doesn’t 53 bits comes out to 17 decimal places). If that’s so, then the 18th, 19th… 25th places should have zeros. Notice in the above output that they have digits other than 0 in them.

Am I misunderstanding something? If so, what?

Answer 1

No, 53 bits means that the 17 decimal places are what you can trust, but because base-10 notation that we use is in a different base from which the double is stored (binary), the later digits are just because 1/2^53 is not exactly 1/10^n, i.e.,

1/2^53 = .0000000000000001110223024625156540423631668090820312500000000