On Wednesday, 24 July 2019 at 16:16:15 UTC, Greatsam4sure wrote:
On Wednesday, 24 July 2019 at 15:57:06 UTC, a11e99z wrote:
On Wednesday, 24 July 2019 at 15:56:13 UTC, a11e99z wrote:
On Wednesday, 24 July 2019 at 15:45:08 UTC, Greatsam4sure
wrote:
int main(){
double mum = 0;
Write("enter a number: ")
readf(" %s\n",&num);
Writeln(num);
}
How do I return the complete number the user enter since I
don't since the user can enter numbers of various length
with dmd approximating it.
I will appreciate any help
readf!" %s\n"( num );
or
num.readf!" %s\n";
or
num = readln.strip.to!double;
writeln(num)
The result is always to six significant figure. I want the
whole number to be diaplay
So, there are several possible issues here. First, floating-point
numbers have limited precision - a double can not correctly
represent all integers above 2^52, for instance. That's 4.5
quadrillion though, so rarely an issue. In your case, as others
have commented, writefln with a correctly chosen format string[1]
will do the trick.
However, what is the correctly chosen format string? This is a
very hard question to answer, as it depends on what number was
typed, and what you mean by 'the complete number the user enter'.
While drug's suggestion of "%.16f" will return an accurate
representation of the floating-point number, it will include a
lot of trailing zeroes that were presumably not present in the
input string. It will also expose floating-point's limitations
when presented with huge numbers, like 1e23 will print
99999999999999991603000.0000000000000000. The best option might
be "%.16g", which will choose the shortest of the scientific
notation and the regular notation, as well as strip trailing
zeroes. Depending on the criteria though, this might still not be
the optimal solution.
--
Simen
1: https://dlang.org/phobos/std_format.html#format-string
