"n00b" <adrienzoltan yahoo.fr> writes:

I tried to use a boolean operator for an array operation :

a[] = b[] < c[];

It compiles but seems to only fill a[] with the result of b[0] < 
c[0].
Is there any "rational" reason to that?
And is there any way to use boolean operator for array operations?

"John Colvin" <john.loughran.colvin gmail.com> writes:

On Thursday, 14 March 2013 at 13:58:45 UTC, n00b wrote:
 I tried to use a boolean operator for an array operation :

 a[] = b[] < c[];

 It compiles but seems to only fill a[] with the result of b[0] 
 < c[0].
 Is there any "rational" reason to that?
 And is there any way to use boolean operator for array 
 operations?

As far as I can tell, this is a bug.

"bearophile" <bearophileHUGS lycos.com> writes:

n00b:

 Is there any "rational" reason to that?

It's not implemented (and it's a bug that it returns something 
different). Take a look in Bugzilla if it's already there.


 And is there any way to use boolean operator for array 
 operations?

I think the only supported boolean vec operation is assignment.

Bye,
bearophile

"bearophile" <bearophileHUGS lycos.com> writes:

 Take a look in Bugzilla if it's already there.

Code for a bug report:


import core.stdc.stdio: printf;
void main() {
     int[3] a = [1, 5, 1];
     int[3] b = [1, 1, 5];
     int[3] c;
     c[] = a[] < b[];
     printf("%d %d %d", c[0], c[1], c[2]);
}


Output:
0 0 0


Expected output:
0 0 1

Bye,
bearophile

"n00b" <adrienzoltan yahoo.fr> writes:

This has already been reported
http://d.puremagic.com/issues/show_bug.cgi?id=5636

"Andrea Fontana" <nospam example.com> writes:

On Thursday, 14 March 2013 at 13:58:45 UTC, n00b wrote:
 I tried to use a boolean operator for an array operation :

 a[] = b[] < c[];

 It compiles but seems to only fill a[] with the result of b[0] 
 < c[0].
 Is there any "rational" reason to that?
 And is there any way to use boolean operator for array 
 operations?

Maybe this way:

int a[] = [1,2,3];
int b[] = [3,2,1];

a.zip(b).map!"a[0]<a[1]".writeln;
return;

"Andrea Fontana" <nospam example.com> writes:

On Thursday, 14 March 2013 at 13:58:45 UTC, n00b wrote:
 I tried to use a boolean operator for an array operation :

 a[] = b[] < c[];

 It compiles but seems to only fill a[] with the result of b[0] 
 < c[0].
 Is there any "rational" reason to that?

Yes i think there is a rational reason. Check this:

int a[] = [1,0,0];
int b[] = [0,1,0];
int c[] = [1,1,0];

bool ab = a[] < b[]; // False.  a[0] > b[0]
bool ac = a[] < c[]; // True.  a[0] == c[0] but a[1] < c[1];

writeln(ab);
writeln(ac);
c[] = ab; // <-- assign!

You see? Your code do this:

a[] = (b[] < c[]);

=?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:

On 03/14/2013 10:45 AM, Andrea Fontana wrote:

 You see? Your code do this:

 a[] = (b[] < c[]);

I thought the same thing at first but note the brackets after b and c. 
Those should make this an array-wise operation.

For all elements of a to be the same value, one would not write the 
brackets:

     a[] = b < c;

Some of the array-wise operations are confusing. :/

Ali

"Andrea Fontana" <nospam example.com> writes:

On Thursday, 14 March 2013 at 17:56:07 UTC, Ali Çehreli wrote:
 On 03/14/2013 10:45 AM, Andrea Fontana wrote:

 You see? Your code do this:

 a[] = (b[] < c[]);

 I thought the same thing at first but note the brackets after b 
 and c. Those should make this an array-wise operation.

 For all elements of a to be the same value, one would not write 
 the brackets:

     a[] = b < c;

 Some of the array-wise operations are confusing. :/

 Ali

Doesn't a[] means copy of a?
So copyofa[] < copyofb[] == bool

I don't think a boolean operator can be array-wise... I always 
think X < Y return a single bool, indipendently from X or Y type 
(class, array or what else). Math operations is a different topic 
in my mind.

"bearophile" <bearophileHUGS lycos.com> writes:

Andrea Fontana:

 I always think X < Y return a single bool,

Having a practice with NumPy and the like, I think X < Y as 
returning as many bools as the length of X and Y:

 from numpy import *
 a = array([1, 5, 1])
 b = array([1, 1, 5])
 a < b



array([False, False,  True], dtype=bool)


Programmers with experience of Matlab, Matcad, etc, feel the same.

Bye,
bearophile

"Andrea Fontana" <nospam example.com> writes:

On Thursday, 14 March 2013 at 18:41:25 UTC, bearophile wrote:
 Andrea Fontana:

 I always think X < Y return a single bool,

 Having a practice with NumPy and the like, I think X < Y as 
 returning as many bools as the length of X and Y:

 from numpy import *
 a = array([1, 5, 1])
 b = array([1, 1, 5])
 a < b



 array([False, False,  True], dtype=bool)


 Programmers with experience of Matlab, Matcad, etc, feel the 
 same.

 Bye,
 bearophile

But does sort!"a<b" still work if it returns an array?

It doesn't make sense for me :) if you ask me if a int vector is
lesser than another one I don't mean every single component but
the whole vector. Same goes for strings or any other range.

"bearophile" <bearophileHUGS lycos.com> writes:

Andrea Fontana:

 But does sort!"a<b" still work if it returns an array?

See the difference in syntax:

a < b
a[] < b[]

Bye,
bearophile

Marco Leise <Marco.Leise gmx.de> writes:

Am Thu, 14 Mar 2013 22:15:11 +0100
schrieb "bearophile" <bearophileHUGS lycos.com>:

 Andrea Fontana:
 
 But does sort!"a<b" still work if it returns an array?

 
 See the difference in syntax:
 
 a < b
 a[] < b[]
 
 Bye,
 bearophile

This is "just" a syntax ambiguity. a[] takes the complete
slice of the array 'a'. And a dynamic array in D is a slice.
So if you use a or a[] in an expression doesn't make much of a
difference.

-- 
Marco

"bearophile" <bearophileHUGS lycos.com> writes:

Marco Leise:

 This is "just" a syntax ambiguity. a[] takes the complete
 slice of the array 'a'. And a dynamic array in D is a slice.
 So if you use a or a[] in an expression doesn't make much of a
 difference.

Yet in D the only accepted syntax to perform a vector op sum is 
to use add square brackets both operands:


void main() {
     auto a = new int[5];
     auto b = new int[5];
     auto c = new int[5];
     c[] = a[] + b[]; // OK
     c[] = a + b; // Error: invalid array operation
     c[] = a[] + b; // Error: invalid array operation
     c[] = a + b[]; // Error: invalid array operation
}


Bye,
bearophile