D: What will s refer to in foreach(ref s; sections[0..$/2+1])

Question

What are the indices these two foreach loops will iterate over:

struct Section{}
Section[] sections;

// assuming we have 10 sections
foreach(ref s; sections[0..$/2+1]) {
    // do something
}       

foreach_reverse(ref s; sections[$/2..$])
{
    // do something
}

Well, I was answering the "what will s refer to..." as the title says. Please change the title if you don't mean to ask that. — Gassa, Apr 05 '14 at 13:24
The first sentence of your question? If that's what you are interested in. — Gassa, Apr 05 '14 at 13:46

ratchet freak · Accepted Answer · 2014-04-05T13:34:09.870

3

the first will iterate over 0-sections.length/2+1=6 (exclusive) and the second will iterate 10-5

the $ in the index refers to the length of the array

and the array[a..b] notation expands to array.opSlice(a,b) that returns a partial view of the array

edited Apr 05 '14 at 13:34

answered Apr 05 '14 at 13:05

ratchet freak

47,288
5
68
106

I get 0..$/2, but what is the effect of the +1? Thanks. – user3477549 Apr 05 '14 at 13:13
Good explanation, thanks. I wonder though how d will behave if length is odd, say 9. Will it be 0-ceil(sections.length/2)+1=6 ? – user3477549 Apr 05 '14 at 13:23
no integer divide will `floor` (more accurately it will cut of the fractional part) – ratchet freak Apr 05 '14 at 13:25
Thanks alot. One thing is disturbing me though.. foreach_reverse(ref s; sections[$/2..$]). Wouldn`t that be 10-5 not 5-10? – user3477549 Apr 05 '14 at 13:29
ah yes I've been a bit hasty in answering it seems – ratchet freak Apr 05 '14 at 13:33

score 1 · Answer 2 · edited May 23 '17 at 12:05

In D, an empty struct has a size 1. The rationale is the same as in C++: see here for a relevant question and the rationale. So, your array will contain dummy 1-byte elements.

Let us check this behavior:

import std.stdio;

struct Section{}
Section[] sections = new Section [10];

void main () {
    foreach(ref s; sections[0..$/2+1]) {
        writeln (&s);
    }
}

To me, this prints:

which are consecutive addresses in memory.

D: What will s refer to in foreach(ref s; sections[0..$/2+1])

2 Answers2