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digitalmars.D - Operate with void[]

reply "Sergey" <httpal gmail.com> writes:
  Hello!

I have the following code:

//--------------------------------
import std.stdio;
import std.file;
import std.array;

int main(){
     void[] str = std.file.read("some_file.txt");
     writeln(str);
     // [32, 32, 55, 56, 9,10]

     // I'm looking for 10 and I change it to 9,9,9,10.
     // and I want to get this [32, 32, 55, 56, 9,9,9,9,10]

     std.file.write("some_file.txt", str2);
     return 0;
}
//--------------------------------

How do I operate with "str" to get the desired result?

Thanks in advance.
Regards, Sergey
Jan 14 2014
parent reply "Adam D. Ruppe" <destructionator gmail.com> writes:
You can't really use void[] directly - cast it to ubyte[] then 
use it and you can write each index by byte.
Jan 14 2014
parent reply "Sergey" <httpal gmail.com> writes:
On Wednesday, 15 January 2014 at 02:56:01 UTC, Adam D. Ruppe 
wrote:

 You can't really use void[] directly - cast it to ubyte[] then 
 use it and you can write each index by byte.


And then back to void[] write to a file?
Jan 14 2014
parent reply "Adam D. Ruppe" <destructionator gmail.com> writes:
On Wednesday, 15 January 2014 at 02:58:50 UTC, Sergey wrote:

 And then back to void[] write to a file?


That's done automatically - any array will implicitly cast to 
const(void)[].

If you're writing a function that can take any kind of data, for 
example, std.file.write, it is nice to take a const void[] since 
then you can pass anything to it:

write("myfile", "a string"); // ok
write("myfile", [0, 1, 2,3]); // ok

and so on.


Going from void[] to any other array requires a cast, since it 
needs to know what kind of data you want to look at (void has no 
meaning itself, so indexing it would be nonsense), but going from 
something else to void happens implicitly.
Jan 14 2014
parent reply "Sergey" <httpal gmail.com> writes:
On Wednesday, 15 January 2014 at 03:11:27 UTC, Adam D. Ruppe 
wrote:

 On Wednesday, 15 January 2014 at 02:58:50 UTC, Sergey wrote:

 And then back to void[] write to a file?


 That's done automatically - any array will implicitly cast to 
 const(void)[].

 If you're writing a function that can take any kind of data, 
 for example, std.file.write, it is nice to take a const void[] 
 since then you can pass anything to it:

 write("myfile", "a string"); // ok
 write("myfile", [0, 1, 2,3]); // ok

 and so on.


 Going from void[] to any other array requires a cast, since it 
 needs to know what kind of data you want to look at (void has 
 no meaning itself, so indexing it would be nonsense), but going 
 from something else to void happens implicitly.


So, please tell me how do I make replace:

//-------------
import std.stdio;
import std.file;
import std.array;


int main(){
	auto fl_txt = "E:/hosp/TMP/file.TXT";
	void[] str = std.file.read(fl_txt);
	ubyte[] pert = cast(ubyte[])str;	

	str2[0] = replace(pert, [9], [9,9,10]); // ?????????	

	std.file.write(fl_txt, pert);
	return 0;
}
//-------------
Jan 14 2014
parent reply "Adam D. Ruppe" <destructionator gmail.com> writes:
On Wednesday, 15 January 2014 at 03:53:18 UTC, Sergey wrote:

 	str2[0] = replace(pert, [9], [9,9,10]); // ?????????	


Try changing that line to this:

         pert = replace(pert, cast(ubyte[]) [9], cast(ubyte[]) 
[9,9,10]);

There's no str2 variable, so I assigned to pert instead. I casted 
the other arrays to ubyte[] because otherwise it throws all kinds 
of errors about ints instead of bytes (number literals are ints 
and the replace function is a bit picky)
Jan 14 2014
parent reply "Sergey" <httpal gmail.com> writes:
On Wednesday, 15 January 2014 at 04:04:05 UTC, Adam D. Ruppe 
wrote:

 On Wednesday, 15 January 2014 at 03:53:18 UTC, Sergey wrote:

 	str2[0] = replace(pert, [9], [9,9,10]); // ?????????	


 Try changing that line to this:

         pert = replace(pert, cast(ubyte[]) [9], cast(ubyte[]) 
 [9,9,10]);

 There's no str2 variable, so I assigned to pert instead. I 
 casted the other arrays to ubyte[] because otherwise it throws 
 all kinds of errors about ints instead of bytes (number 
 literals are ints and the replace function is a bit picky)


Sorry, str2 is my mistake.
And I think the last question:
when the number of bytes with the number 9 is not known how to 
attach them?


pert = replace(pert, cast(ubyte[]) [9], cast(ubyte[])[9] +
ubyte[])[9] + ...n... + ubyte[])[10]);
Jan 14 2014
parent "Sergey" <httpal gmail.com> writes:
On Wednesday, 15 January 2014 at 05:21:26 UTC, Sergey wrote:

 On Wednesday, 15 January 2014 at 04:04:05 UTC, Adam D. Ruppe 
 wrote:

 On Wednesday, 15 January 2014 at 03:53:18 UTC, Sergey wrote:

 	str2[0] = replace(pert, [9], [9,9,10]); // ?????????	


 Try changing that line to this:

        pert = replace(pert, cast(ubyte[]) [9], cast(ubyte[]) 
 [9,9,10]);

 There's no str2 variable, so I assigned to pert instead. I 
 casted the other arrays to ubyte[] because otherwise it throws 
 all kinds of errors about ints instead of bytes (number 
 literals are ints and the replace function is a bit picky)


 Sorry, str2 is my mistake.
 And I think the last question:
 when the number of bytes with the number 9 is not known how to 
 attach them?


 pert = replace(pert, cast(ubyte[]) [9], cast(ubyte[])[9] +
 ubyte[])[9] + ...n... + ubyte[])[10]);


again sorry....:)))
some_ubyte = some_ubyte ~ cast(ubyte[])[9]
Jan 14 2014