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!

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.

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.

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].

That's a really informative response.  Thank you!