• Jonathan Levi (35/35) Nov 23 2019 I have a `double` (from a math equation) which when logged is
• Timon Gehr (6/13) Nov 23 2019 You can print the number at a higher precision:
• Jonathan Levi (3/8) Nov 23 2019 Oh, right, duh, thanks.

Jonathan Levi <catanscout gmail.com> writes:

I have a `double` (from a math equation) which when logged is 
`-1` but when a `acos` it it returns `-nan`?

The `double` has different bit representation than a normal `-1` 
but I do not see how that is messing up its `acos`.

The `double`'s representation is: true,1023,58720256.

This will reproduce the problem:

```
import std;

void dWrite(double d) {
     import std.stdio;
     import std.conv;
     import std.bitmanip;
     auto dr = DoubleRep(d);
     writeln(d,"\t",dr.sign,"\t", dr.exponent,"\t", dr.fraction);
}
double fromRep(bool sign, ushort exponent, ulong fraction) {
     DoubleRep r;
     r.sign = sign;
     r.exponent = exponent;
     r.fraction = fraction;
     return r.value;
}

void main() {
     double failing = fromRep(true,1023,58720256);
     double lookalike = -1;

     failing.dWrite;
     lookalike.dWrite;

     failing.acos.dWrite;
     lookalike.acos.dWrite;
}
```

Any idea why that is and how I could solve it?

BTW this is the how that number is created: 
`cos(a.angle)*cos(b.angle) - 
dot(a.axis*sin(a.angle),(b.axis*sin(b.angle)))`

Timon Gehr <timon.gehr gmx.ch> writes:

On 23.11.19 18:31, Jonathan Levi wrote:
 ...
 Any idea why that is and how I could solve it?
 
 BTW this is the how that number is created: `cos(a.angle)*cos(b.angle) - 
 dot(a.axis*sin(a.angle),(b.axis*sin(b.angle)))`
 
 

You can print the number at a higher precision:

writefln!"%.16f"(failing); // -1.0000000130385160

I.e., it is actually slightly smaller than -1.
You either have to debug your data source or manually clip the value 
into the range [-1.0,1.0] before you call acos.

Jonathan Levi <catanscout gmail.com> writes:

On Saturday, 23 November 2019 at 18:09:43 UTC, Timon Gehr wrote:
 You can print the number at a higher precision:

 writefln!"%.16f"(failing); // -1.0000000130385160

Oh thanks, that is how that can be done.

 I.e., it is actually slightly smaller than -1.
 You either have to debug your data source or manually clip the 
 value into the range [-1.0,1.0] before you call acos.

Oh, right, duh, thanks.