• Darren (13/13) Jul 31 2016 Hey, all.
• Seb (6/20) Jul 31 2016 Ehm how do you you want to represent 1_000 in one dchar?
• Johannes Loher (16/42) Jul 31 2016 An immutable array of dchars is a dstring, not a string (which is an
• ag0aep6g (13/16) Jul 31 2016 Because it gives you a dchar with the numeric value 5 which is some
• Darren (1/1) Jul 31 2016 That's a really informative response. Thank you!

Darren <darren-x hotmail.co.uk> writes:

Hey, all.

I'm pretty much a programming novice, so I hope you can bear with 
me.  Does anyone know how I can change an int into a char 
equivalent?

e.g.
int i = 5;
dchar value;
?????
assert(value == '5');

If I try and cast it to dchar, I get messed up output, and I'm 
not sure how to use toChars (if that can accomplish this).

I can copy+paste the little exercise I'm working on if that helps?

Thanks in advance!

Seb <seb wilzba.ch> writes:

On Sunday, 31 July 2016 at 21:31:52 UTC, Darren wrote:
 Hey, all.

 I'm pretty much a programming novice, so I hope you can bear 
 with me.  Does anyone know how I can change an int into a char 
 equivalent?

 e.g.
 int i = 5;
 dchar value;
 ?????
 assert(value == '5');

 If I try and cast it to dchar, I get messed up output, and I'm 
 not sure how to use toChars (if that can accomplish this).

 I can copy+paste the little exercise I'm working on if that 
 helps?

 Thanks in advance!

Ehm how do you you want to represent 1_000 in one dchar?
You need to format it, like here.

     import std.format : format;
     assert("%d".format(1_000) == "1000");

Note that you get an array of dchars (=string), not a single one.

Johannes Loher <johannesloher fg4f.de> writes:

Am 31.07.2016 um 23:46 schrieb Seb:
 On Sunday, 31 July 2016 at 21:31:52 UTC, Darren wrote:
 Hey, all.

 I'm pretty much a programming novice, so I hope you can bear with me. 
 Does anyone know how I can change an int into a char equivalent?

 e.g.
 int i = 5;
 dchar value;
 ?????
 assert(value == '5');

 If I try and cast it to dchar, I get messed up output, and I'm not
 sure how to use toChars (if that can accomplish this).

 I can copy+paste the little exercise I'm working on if that helps?

 Thanks in advance!

 
 Ehm how do you you want to represent 1_000 in one dchar?
 You need to format it, like here.
 
     import std.format : format;
     assert("%d".format(1_000) == "1000");
 
 Note that you get an array of dchars (=string), not a single one.

An immutable array of dchars is a dstring, not a string (which is an
immutable array of chars). It is true however, that you should not
convert to dchar, but to string (or dstring, if you want utf32, but i
see no real reason for this, if you are only dealing with numbers),
because of the reason mentioned above. Another solution for this would
be using "to":

import std.conv : to;

void main()
{
    int i = 5;
    string value = i.to!string;
    assert(value == "5");
}

If you know that your int only has one digit, and you really want to get
it as char, you can always use value[0].

ag0aep6g <anonymous example.com> writes:

On 07/31/2016 11:31 PM, Darren wrote:
 If I try and cast it to dchar, I get messed up output,

Because it gives you a dchar with the numeric value 5 which is some 
control character.

 and I'm not sure
 how to use toChars (if that can accomplish this).

     value = i.toChars.front;

toChars converts the number to a range of chars. front takes the first 
of them.

Similarly, you could also convert to a (d)string and take the first 
character:

     value = i.to!dstring[0];

Or if you want to appear clever, add i to '0':

     value = '0' + i;

I'd generally prefer toChars.front here. to!dstring[0] makes an 
allocation you don't need, and '0' + i is more obscure and bug-prone.

Darren <darren-x hotmail.co.uk> writes:

That's a really informative response.  Thank you!