• Barry Denton (8/8) Apr 14 2008 from the examples I try to do
• Jarrett Billingsley (23/31) Apr 14 2008 You can't declare a statically-sized array with a non-constant value. I...
• Barry Denton (2/54) Apr 14 2008 Thanks !

Barry Denton <basse42 yahoo.com> writes:

from the examples I  try to do 

int COUNT = 10000;
    char [COUNT][] itemStrings ;
    for (int i = 0; i < COUNT; i++) {
      
      itemStrings [i][] = ("item " ~ to!(char[]) (i));

2 problems -does not accept COUNT as the size
and if I use char[10000] its OK but after compiles OK but index out of bounds
error given on itemStrings[][]   line

"Jarrett Billingsley" <kb3ctd2 yahoo.com> writes:

"Barry Denton" <basse42 yahoo.com> wrote in message 
news:fu0hfh$29mq$1 digitalmars.com...
 from the examples I  try to do

 int COUNT = 10000;
    char [COUNT][] itemStrings ;
    for (int i = 0; i < COUNT; i++) {

      itemStrings [i][] = ("item " ~ to!(char[]) (i));

 2 problems -does not accept COUNT as the size

You can't declare a statically-sized array with a non-constant value.  If 
you know you will always have 10000 values, use:

const int COUNT = 10000;

If the number of values can vary at runtime, use a dynamically allocated 
array instead:

char[][] itemStrings = new char[][](n, count);

 and if I use char[10000] its OK but after compiles OK but index out of 
 bounds error given on itemStrings[][]   line

You haven't given any size to the second dimension of the array.  You've 
declared the array reference but haven't actually created an array.

Maybe you're getting the syntax confused.  The C-style syntax:

char itemStrings[COUNT][];

becomes:

char[][COUNT] itemStrings;

in D.  Your array:

char[COUNT][] itemStrings;

reads as "a dynamic array of statically-sized arrays of length COUNT of 
int".  You read right-to-left.  It seems you want the other way, 
char[][COUNT].

Even then, your loop is still wrong.  You are trying to slice-assign an 
array that doesn't exist.  Simply take out the slice on the left-hand-side:

itemStrings[i] = "item " ~ to!(char([])(i);

And it will fill in itemStrings[i] with the new string.

Barry Denton <basse42 yahoo.com> writes:

Jarrett Billingsley Wrote:

 "Barry Denton" <basse42 yahoo.com> wrote in message 
 news:fu0hfh$29mq$1 digitalmars.com...
 from the examples I  try to do

 int COUNT = 10000;
    char [COUNT][] itemStrings ;
    for (int i = 0; i < COUNT; i++) {

      itemStrings [i][] = ("item " ~ to!(char[]) (i));

 2 problems -does not accept COUNT as the size

 
 You can't declare a statically-sized array with a non-constant value.  If 
 you know you will always have 10000 values, use:
 
 const int COUNT = 10000;
 
 If the number of values can vary at runtime, use a dynamically allocated 
 array instead:
 
 char[][] itemStrings = new char[][](n, count);
 
 and if I use char[10000] its OK but after compiles OK but index out of 
 bounds error given on itemStrings[][]   line

 
 You haven't given any size to the second dimension of the array.  You've 
 declared the array reference but haven't actually created an array.
 
 Maybe you're getting the syntax confused.  The C-style syntax:
 
 char itemStrings[COUNT][];
 
 becomes:
 
 char[][COUNT] itemStrings;
 
 in D.  Your array:
 
 char[COUNT][] itemStrings;
 
 reads as "a dynamic array of statically-sized arrays of length COUNT of 
 int".  You read right-to-left.  It seems you want the other way, 
 char[][COUNT].
 
 Even then, your loop is still wrong.  You are trying to slice-assign an 
 array that doesn't exist.  Simply take out the slice on the left-hand-side:
 
 itemStrings[i] = "item " ~ to!(char([])(i);
 
 And it will fill in itemStrings[i] with the new string.
 
 

Thanks !