On 1/31/18 6:19 PM, Azi Hassan wrote:
On Saturday, 27 January 2018 at 14:13:49 UTC, kdevel wrote:
I would expect this code
enforce3.d
---
import std.exception;
void main ()
{
int i = int.min;
enforce (i > 0);
}
---
to throw an "Enforcement failed" exception, but it doesn't:
$ dmd enforce3.d
$ ./enforce3
[nothing]
I wonder if it's caused by a comparison between signed and unsigned
integers.
No, the answer is, there's a shortcut optimization used by the compiler.
See the discussion elsewhere in this thread.
import std.stdio;
void main ()
{
int zero = 0;
writeln(int.min > 0u);
writeln(int.min > zero);
}
Note that comparing the literal int.min will get folded into a constant,
and do the right thing. You have to assign it a variable to see the
incorrect behavior:
int i = int.min;
writeln(i > 0);
-Steve
