Re: A floating-point puzzle
Jarrett Billingsley wrote: On Wed, Aug 5, 2009 at 10:16 PM, Donnos...@nospam.com wrote: Lars T. Kyllingstad wrote: Lars T. Kyllingstad wrote: Here's a puzzle for you floating-point wizards out there. I have to translate the following snippet of FORTRAN code to D: REAL B,Q,T C -- C |*** COMPUTE MACHINE BASE ***| C -- T = 1. 10T = T + T IF ( (1.+T)-T .EQ. 1. ) GOTO 10 B = 0. 20B = B + 1 IF ( T+B .EQ. T ) GOTO 20 IF ( T+2.*B .GT. T+B ) GOTO 30 B = B + B 30Q = ALOG(B) Q = .5/Q Of course I could just do a direct translation, but I have a hunch that T, B, and Q can be expressed in terms of real.epsilon, real.min and so forth. I have no idea how, though. Any ideas? (I am especially puzzled by the line after l.20. How can this test ever be true? Is the fact that the 1 in l.20 is an integer literal significant?) -Lars I finally solved the puzzle by digging through ancient scientific papers, as well as some old FORTRAN and ALGOL code, and the solution turned out to be an interesting piece of computer history trivia. After the above code has finished, the variable B contains the radix of the computer's numerical system. Perhaps the comment should have tipped me off, but I had no idea that computers had ever been anything but binary. But apparently, back in the 50s and 60s there were computers that used the decimal and hexadecimal systems as well. Instead of just power on/off, they had 10 or 16 separate voltage levels to differentiate between bit values. Not quite. They just used exponents which were powers of 10 or 16, rather than 2. BTW, T == 1/real.epsilon. I don't know what ALOG does, so I've no idea what Q is. Apparently ALOG is just an old name for LOG. At least that's what Google tells me. Then Q is 0.5*ln(0.5). Dunno what use that is.
Re: At compile time
Jarrett Billingsley wrote: *On Wed, Aug 5, 2009 at 7:23 AM, Ary Borenszweiga...@esperanto.org.ar wrote: bearophile escribió: Jarrett Billingsley: C++ has static initialization that occurs before main() too. It's just.. hidden. I see. I have to learn more about C++. Thank you. -- Lars T. Kyllingstad: This is good news! The restrictions you are referring to, are they any of the ones documented here: http://www.digitalmars.com/d/2.0/function.html#interpretation That list has a point regarding what I was trying to do: 4. the function may not be a non-static member, i.e. it may not have a this pointer It's true regarding stucts used as values too, and not just classes... It would be nice if the compiler could say so: I can't evaluate it because you are using a this pointer. With other words, but much more useful than non-constant expression. Yes, oh my God, this is the main reason I don't use CTFE: debugging them is virtually impossible, and the compiler does nothing to help there. Nice error messages for CTFE assignment statements are in the next release. That's probably 50% of the instances of non-constant expression.
noob question
hi im noob here and try learn D language, i try make this example but when run
have 2 problems.
1) when i show name and lastname show me in two lines and i wanna make in the
same line.
2) can´t validate option si ( yes ) or no ( no ) and always show the else line.
what happend?
thx
--
code
--
import std.stdio;
import std.string;
import std.c.stdio;
void main()
{
string _nombre, _apellido, _respuesta;
int _edad, _año, _nacimiento;
_año = 2009;
writefln (Escribe tu nombre: );
_nombre = readln();
writefln (Escribe tu apellido: );
_apellido = readln();
writefln (Hola %s , _nombre ~= _apellido);
writefln (Ingresa el año de tu nacimiento: );
scanf (%d, _nacimiento);
_edad = _año - _nacimiento;
writefln (Tu edad es %d? \t SI \t NO, _edad);
_respuesta = readln();
if ( _respuesta == si)
writeln (Muchas gracias %s, _nombre );
else
writeln (Lo siento %s entonces tienes %d, _nombre, _edad - 1 );
}
how does range.put work
Hello, could someone plz clearify what the exact semantics of put are ? Put works with an appender, but gives me a runtime exception when using an array. Best regards, Oliver
Re: noob question
Reply to llltattoolll, writefln (Your age is %d? \t YES \t NO, _age ); /* here is my second problem, can´t into the response and program*/ _response = readln(); /* show me else line always */ // if all else fails, what are you getting writef(%s\n, cast(ubyte[])chomp(_response)); // might it be a case error? YES vs. yes? if ( chomp(_response) == yes) writefln (thank you %s, _name ); else writefln (Sorry %s you have %d, _name, _age - 1 ); /* this line always show me because can´t validate _response */
Linking Tango + QtD under Ubuntu (with Rebuild)
I'm trying to evaluate QtD. First step is to link something against Tango and Qt, but unfortunately I can't get it to work: rebuild main.d -debug -full -oqobj/ -I~/coding -llqtdcore -llqtdgui - llQtCore -llQtGui I have Tango installed via apt-get (and the Tango repository), as well as libqt4-dev and all its dependencies, but still the linker can resolve almost nothing. When I remove the QtD imports and simply link against Tango (with a test Stdout), it links and runs. Has anybody run into some similar problem? I've never been good at this linking business and I'm out of ideas :( so help would be much appreciated. Regards, Mike
Re: Shared Object with DMD v2.031
On Thu, 06 Aug 2009 04:24:35 +0400, Sergey Gromov wrote: Wed, 5 Aug 2009 20:46:53 + (UTC), teo wrote: On Tue, 04 Aug 2009 18:41:50 +0400, Sergey Gromov wrote: Sun, 2 Aug 2009 11:18:24 + (UTC), teo wrote: On Sun, 02 Aug 2009 02:18:28 +0400, Sergey Gromov wrote: My guess is that test.di is exactly the same as test.d because all the functions are small. Therefore compiling 'dmd prog.d test.di' resolves all symbols statically and the .so is simply not linked. Does your 'prog' have any unresolved symbols in its symbol table? It looks like this is the case. I found following in the symbols: 0804b8cc T _D4test6testMeFZi 0804b8d8 T _D4test9testClassFiZC4test4Test The whole nm output is too long, but if you want I can send it to you. When I use the *--undefined-only* option I get: $ nm -u prog [snip] All unresolved symbols seem to be in glibc. Yes. I've filed a bug: http://d.puremagic.com/issues/show_bug.cgi?id=3226 I don't have the warning you mention. Try to run objdump -r test.o It'll print relocation table for the object file. In my case almost all of them are of type R_386_32 which are load-time relocations. A position-independent code should not contain them. Correct. This is exactly the case. And my environment is identical to yours: $ dmd | head -n1 Digital Mars D Compiler v2.031 $ gcc --version | head -n1 gcc (Ubuntu 4.3.3-5ubuntu4) 4.3.3 $ uname -a Linux dev-phobos 2.6.28-14-server #47-Ubuntu SMP Sat Jul 25 01:18:34 UTC 2009 i686 GNU/Linux So it is reproducible with DMD2 as well.
Re: noob question
Thu, 06 Aug 2009 10:56:33 -0400, lllTattoolll wrote:
hi Lars and thx for help. I fix a little the first problem, now I paste a
verion in english whit coment because i´m lost here.
the example is the same only this is in english for your compresion of my
problem.
--
code
-
import std.stdio;
import std.string;
import std.c.stdio;
void main()
{
string _name, _lastname, _response;
int _age, _year, _birth;
_year = 2009;
writef (Name: );
_name = readln().chomp;
writef (Lastname: );
_lastname = readln().chomp;
writefln (Hello %s , _name, _lastname ); /* here is my first
problem, dont show me the last name or show me in
two
lines ( this fix whit .chomp ) */
The format string must be Hello %s %s to display both operands. Or
you can use writeln instead:
writeln(Hello , _name, , _lastname);
writefln (Year of birth: );
scanf (%d, _birth);
You use scanf here. Scanf only reads the number, and leaves the Enter
symbol in the input stream. When you later call readln() it sees that
Enter from the previous question and exits immediately. You better
use readln() everywhere:
_birth = std.conv.to!int(readln().chomp);
_age = _year - _birth;
writefln (Your age is %d? \t YES \t NO, _age ); /* here is my second
problem, can´t into the response and program*/
_response = readln();/* show me
else line always */
if ( chomp(_response) == yes)
writefln (thank you %s, _name );
else
writefln (Sorry %s you have %d, _name, _age - 1 ); /* this line
always show me because can´t validate _response */
}
The rest seems to work correctly.
Re: how does range.put work
O.K. wrote:
Hello,
could someone plz clearify what the exact semantics of put
are ?
Put works with an appender, but gives me a runtime exception
when using an array.
Best regards,
Oliver
The source code for the standard library comes with the compiler.
If you look in std\array.d, you find this around line 279 (reflowed for
readability):
void put(T, E)(ref T[] a, E e) {
assert(a.length);
a[0] = e; a = a[1 .. $];
}
