• Gary Willoughby (9/9) Jan 01 2014 I'm calling an external C function which returns a string
• Adam D. Ruppe (7/10) Jan 01 2014 you can then do
• Gary Willoughby (2/13) Jan 02 2014 to!(string) works great thanks!
• Gary Willoughby (2/13) Jan 02 2014 Another question: how do i convert const(char)** to string[]?
• bearophile (8/9) Jan 02 2014 If you know that you have N strings, then a solution is
• Adam D. Ruppe (4/6) Jan 02 2014 Or if it is zero terminated, maybe
• monarch_dodra (2/8) Jan 02 2014 0-terminated arrays of non-strings :puke:
• Gary Willoughby (5/14) Jan 02 2014 Thanks, both work well:
• bearophile (11/14) Jan 02 2014 This is a feature that works with all pointers to a sequence of
• uc (2/2) Jan 02 2014 You are going to need the length of your c char*[] then a
• uc (2/2) Jan 02 2014 i'll answer in code
• John Colvin (7/12) Jan 01 2014 You're not asking to print the string, by dereferencing the

"Gary Willoughby" <dev nomad.so> writes:

I'm calling an external C function which returns a string 
delivered via a char*. When i print this string out, like this:

char* result = func();

writefln("String: %s", *result);

I only get one character printed. I guess this is expected 
because i'm only returned a pointer to the first char. Instead of 
incrementing the pointer until i reach a null is there an easy 
way to convert this to a D string or get writeln to print the 
entire char array?

"Adam D. Ruppe" <destructionator gmail.com> writes:

On Wednesday, 1 January 2014 at 23:03:06 UTC, Gary Willoughby 
wrote:
 I'm calling an external C function which returns a string 
 delivered via a char*. When i print this string out, like this:

 char* result = func();'

you can then do

string r = to!string(result);

or

char[] r = result[0 .. strlen(result)];


and use that/

"Gary Willoughby" <dev nomad.so> writes:

On Wednesday, 1 January 2014 at 23:09:05 UTC, Adam D. Ruppe wrote:
 On Wednesday, 1 January 2014 at 23:03:06 UTC, Gary Willoughby 
 wrote:
 I'm calling an external C function which returns a string 
 delivered via a char*. When i print this string out, like this:

 char* result = func();'

 you can then do

 string r = to!string(result);

 or

 char[] r = result[0 .. strlen(result)];


 and use that/

to!(string) works great thanks!

"Gary Willoughby" <dev nomad.so> writes:

On Wednesday, 1 January 2014 at 23:09:05 UTC, Adam D. Ruppe wrote:
 On Wednesday, 1 January 2014 at 23:03:06 UTC, Gary Willoughby 
 wrote:
 I'm calling an external C function which returns a string 
 delivered via a char*. When i print this string out, like this:

 char* result = func();'

 you can then do

 string r = to!string(result);

 or

 char[] r = result[0 .. strlen(result)];


 and use that/

Another question: how do i convert const(char)** to string[]?

"bearophile" <bearophileHUGS lycos.com> writes:

Gary Willoughby:

 Another question: how do i convert const(char)** to string[]?

If you know that you have N strings, then a solution is 
(untested):

pp[0 .. N].map!text.array

If it doesn't work, try:

pp[0 .. N].map!(to!string).array

Bye,
bearophile

"Adam D. Ruppe" <destructionator gmail.com> writes:

On Thursday, 2 January 2014 at 15:31:25 UTC, bearophile wrote:
 If you know that you have N strings, then a solution is 
 (untested):

Or if it is zero terminated, maybe

pp.until!"a is null".map!text.array


Though personally I'd just use the plain old for loop.

"monarch_dodra" <monarchdodra gmail.com> writes:

On Thursday, 2 January 2014 at 15:53:40 UTC, Adam D. Ruppe wrote:
 On Thursday, 2 January 2014 at 15:31:25 UTC, bearophile wrote:
 If you know that you have N strings, then a solution is 
 (untested):

 Or if it is zero terminated, maybe

 pp.until!"a is null".map!text.array


 Though personally I'd just use the plain old for loop.

0-terminated arrays of non-strings :puke:

"Gary Willoughby" <dev nomad.so> writes:

On Thursday, 2 January 2014 at 15:31:25 UTC, bearophile wrote:
 Gary Willoughby:

 Another question: how do i convert const(char)** to string[]?

 If you know that you have N strings, then a solution is 
 (untested):

 pp[0 .. N].map!text.array

 If it doesn't work, try:

 pp[0 .. N].map!(to!string).array

 Bye,
 bearophile

Thanks, both work well:

I've noticed that const(char)** can be accessed via indexes:

writefln("%s", pp[0].to!(string)); //etc.

cool!

"bearophile" <bearophileHUGS lycos.com> writes:

Gary Willoughby:

 I've noticed that const(char)** can be accessed via indexes:

 writefln("%s", pp[0].to!(string)); //etc.

 cool!

This is a feature that works with all pointers to a sequence of 
items, like in C. But array bounds are not verified, so it's more 
bug-prone. So if you know the length it's better to slice the 
pointer as soon as possible, and then use the slice:

auto ap = pp[0 .. N];
writefln("%s", ap[0].text);

Or just:

printf("%s\n", ap[0]);

Bye,
bearophile

"uc" <uc uc.uc> writes:

You are going to need the length of your c char*[] then a 
for-loop should do it :D

"uc" <uc uc.uc> writes:

i'll answer in code
http://dpaste.dzfl.pl/2bb1a1a8

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

On Wednesday, 1 January 2014 at 23:03:06 UTC, Gary Willoughby 
wrote:
 I'm calling an external C function which returns a string 
 delivered via a char*. When i print this string out, like this:

 char* result = func();

 writefln("String: %s", *result);

 I only get one character printed.

You're not asking to print the string, by dereferencing the 
pointer you're literally asking to print the first character.

If you don't want to have to walk the length of the the string to 
make the D array equivalent, you could always wrap it in a 
forward range to emulate c string handling.