You'd just have to convert each element of the
array separately.
I found the formatting options, they are in std.format..
(I was apparently searching for the 'g' option.. )
this seems to work (using std2.string.format)
double[] d1;
d1=[double.max,double.min];
char[] c;
c=format(
On Sat, Jul 11, 2009 at 5:50 PM, Saaaem...@needmail.com wrote:
double d[2] = [ 0, 1, double.max];
char[] c = format(d);
How do I get c to represent full precision?
[0,1,1.7976931348623157e+308] // but then with double.max being
represented fully
You want a 309-digit number consisting
double d[2] = [ 0, 1, double.max];
char[] c = format(d);
How do I get c to represent full precision?
[0,1,1.7976931348623157e+308] // but then with double.max being
represented fully
You want a 309-digit number consisting mostly of 0s?
Yes, but only if they are necessary.
0 doesn't need
On Sat, Jul 11, 2009 at 6:44 PM, Saaaem...@needmail.com wrote:
double d[2] = [ 0, 1, double.max];
char[] c = format(d);
How do I get c to represent full precision?
[0,1,1.7976931348623157e+308] // but then with double.max being
represented fully
You want a 309-digit number consisting
On Sat, Jul 11, 2009 at 7:39 PM, Saaaem...@needmail.com wrote:
Um, doubles don't have infinite precision. See those digits that it
output? That's all you get. Those are the only digits that are
necessary because those are the only digits that are *stored*. Just
because it's followed by
