Bus error accessing char [] by index - OS X

[Regan Heath]

Regan Heath wrote:
> Regan Heath wrote:
>> Cam MacRae wrote:
>>> BCS Wrote:
>>>
>>>> Reply to Cam,
>>>>
>>>>
>>>> On linux, string literals are in read only memory space. As for OS
>>>> X...???
>>>>
>>>>
>>>
>>> Thanks mate - that could be it. I was just following along the D
>>> Transition guide (http://en.wikibooks.org/wiki/D_Transition_Guide).
>>> Perhaps it's out of date...
>>
>> More like "windows centric" as string literals can be written to on 
>> windows.
>>
>> Simple fix:
>>
>>         char[] a = firstname.dup, b = firstname;
>>         a[0] = 'H'; //5
> 
> Actually, this breaks the example entirely as b is no longer modified by 
> a[0] - 'H' oops.

Here is my suggested fix - I don't have DMD 1.0 to test this with so 
someone else is going to have to edit the wiki (after testing this works!)

import std.stdio;

  void main() {
         char[] firstname, lastname, fullname; //1
         firstname = "Walter".dup; //2
         lastname = "Bright"; //3
         fullname = firstname ~ " " ~ lastname; //4
         writefln("Congratulations on making a great language " ~ 
fullname); //5

         char[] a = firstname, b = firstname;
         a[0] = 'H'; //6
         writefln(b); //prints "Halter"
         writefln("Your name is still %s, right?", fullname); //7, 
prints Walter Bright
  }

    1. Strings are nothing more than character arrays. You'll see more 
about arrays later, but for now, know that character arrays are not a 
special case. The one major note is that D strings are not null 
terminated. Arrays simply keep track of their length.
    2. dup used here to create a copy of the string "Walter" (this is 
because string literals are read-only on some OS's)
    3. No strcpy here. In fact, this is more like reassigning a pointer 
(ie, char *lastname = "Bright").
    4. ~ is the concatenation operator. There is no ambiguity between + 
and ~ for strings.
    5. One way of outputting, although number 7 is better
    6. Since a is really a pointer to firstname, firstname (and hence b) 
actually get modified on this line.
    7. But since fullname was created through concatenation, it remains 
unchanged.