• Hugo Florentino (46/46) Mar 08 2012 Hello,
• =?UTF-8?B?QWxpIMOHZWhyZWxp?= (30/53) Mar 08 2012 One of many ways:

Hugo Florentino <sysadmin cips.cu> writes:

Hello,

I am a learning D, and so far I am enjoying it. However, since I have a 
few doubts I cannot clear just by reading the available documentation, I 
registered in this list, so please bear with me if my questions sound a 
bit dumb.

So my first doubt is this:

What is the proper D syntax to initialize a static array with 
consecutive integers starting with 1?

I know I could simply use "for" like in C, but while reading the web 
documentation on arrays, I noticed the vector notation. According to the 
documentation, this code:
T[] a, b;
...
a[] = b[] + 4;

is equivalent to this code:
T[] a, b;
...
for (size_t i = 0; i < a.length; i++) a[i] = b[i] + 4;

Now, suppose I declare a static array:
int[100] myarray;

What would be the substitution in vector notation for this code?
for (int i =0; i < myarray.length; i++) myarray[i] = i +1;

I have tried these, and neither seems to work as expected (they simply 
assign 0+1 to every element):
myarray[] += 1;
myarray[] = myarray[] + 1;

I also tried using the array as an aggregate in a foreach statement:
foreach(int i, int j, myarray) j = i + 1;

However, it does not work this way because apparently the j variable 
seems to work only for reading, not for assigning. I wonder why this 
limitation in behavior, if according to the documentation:

"If there are two variables declared, the first is said to be the index 
and the second is said to be the value [set to the elements of the 
array, one by one]"

So if j refers to the value, the intuitive thing IMHO would be assigning 
to each element when one assigns to j. Oherwise one would have to do 
something redundant like this:
foreach(int i, int j, myarray) myarray[i] = j = i + 1;

Or maybe this (which renders j useless altogether)
foreach(int i, int j, myarray) myarray[i] = i + 1;

Or even something like this (which works, but misses the advantage of 
using the array as an aggregate):
foreach(int i; 0..myarray.length) myarray[i] = i + 1;

Or I am missing something?
Either way, please advice the recommended D way.

Regards, Hugo

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

On 03/08/2012 06:02 PM, Hugo Florentino wrote:

 What is the proper D syntax to initialize a static array with
 consecutive integers starting with 1?

One of many ways:

int[N] makeArray(size_t N)()
{
     int[N] result;

     foreach (int i, ref element; result) {
         element = i;
     }

     return result;
}

void main()
{
     assert(makeArray!4() == [ 0, 1, 2, 3 ]);
}

 I know I could simply use "for" like in C, but while reading the web
 documentation on arrays, I noticed the vector notation. According to the
 documentation, this code:
 T[] a, b;
 ...
 a[] = b[] + 4;

 is equivalent to this code:
 T[] a, b;
 ...
 for (size_t i = 0; i < a.length; i++) a[i] = b[i] + 4;

It should be possible to use a special type and keep state to achieve it 
with the arraywise syntax but this task is natural for that syntax.

 I also tried using the array as an aggregate in a foreach statement:
 foreach(int i, int j, myarray) j = i + 1;

 However, it does not work this way because apparently the j variable
 seems to work only for reading, not for assigning. I wonder why this
 limitation in behavior, if according to the documentation:

 "If there are two variables declared, the first is said to be the index
 and the second is said to be the value [set to the elements of the
 array, one by one]"

 So if j refers to the value,

That's the problem. j does not refer to the element, it is a copy of it. 
You must use the 'ref' keyword as in the code that I have shown above.

 Either way, please advice the recommended D way.

There are many ways of initializing a fixed-length array with 
consecutive integers. Here is another one:

import std.algorithm;
import std.range;

void main()
{
     int[4] array = array(iota(4));
     assert(array == [ 0, 1, 2, 3 ]);
}

 Regards, Hugo

Ali

P.S. There is also the D.learn newsgroup where such threads are very 
welcome at. :)