Hi guys! I had this homework assignment for data structures that has a pretty easy solution in C++. Reading input like this...

1 2 3 # $
4 3 * ! #
20 3 / # $ #
62 # $
2 3 8 * + #
4 48 4 2 + / #
SUM # $
1 2 3 4 5 #
R #
@

...where "@" denotes the end of input is fairly simple in C++:

string token = "";
while (token != "@") {
  //handle input
}

Note that having newlines doesn't matter at all; every token is just assumed to be separated by "whitespace". However in D, I looked around could not find a solution better than this:

foreach (line; stdin.byLine) {
  foreach (token; line.split) {
    //handle input
  }
}

Is there any way to do this without two loops/creating an array? "readf(" %d", &token);" wasn't cutting it either.

Thanks.

I think this should work:

import std.stdio;

void main() {
        string token;
        while(readf("%s ", &token))
                writeln(token);
}


Have you tried that? What is wrong with it if you have?

On Tuesday, 21 April 2015 at 01:46:53 UTC, Adam D. Ruppe wrote:
> I think this should work:
>
> import std.stdio;
>
> void main() {
>         string token;
>         while(readf("%s ", &token))
>                 writeln(token);
> }
>
>
> Have you tried that? What is wrong with it if you have?

It'll just leave some trailing whitespace, which I don't want. And somehow doing token = token.stip STILL leaves that whitespace somehow...

On Tuesday, 21 April 2015 at 02:04:24 UTC, TheGag96 wrote:
> It'll just leave some trailing whitespace, which I don't want.

oh it also keeps the newlines attached. Blargh.

Well, forget the D functions, just use the C functions:

import core.stdc.stdio;

void main() {
        char[16] token;
        while(scanf("%15s", &token) != EOF)
                printf("**%s**\n", token.ptr);
}



You could convert to a string if needed with import std.conv; to!string(token.ptr), but if you can avoid that, you should, this loop has no allocations which is a nice thing.

On Tuesday, 21 April 2015 at 01:31:58 UTC, TheGag96 wrote:
> Hi guys! I had this homework assignment for data structures that has a pretty easy solution in C++. Reading input like this...
>
> 1 2 3 # $
> 4 3 * ! #
> 20 3 / # $ #
> 62 # $
> 2 3 8 * + #
> 4 48 4 2 + / #
> SUM # $
> 1 2 3 4 5 #
> R #
> @
>
> ...where "@" denotes the end of input is fairly simple in C++:
>
> string token = "";
> while (token != "@") {
>   //handle input
> }
>
> Note that having newlines doesn't matter at all; every token is just assumed to be separated by "whitespace". However in D, I looked around could not find a solution better than this:
>
> foreach (line; stdin.byLine) {
>   foreach (token; line.split) {
>     //handle input
>   }
> }
>
> Is there any way to do this without two loops/creating an array? "readf(" %d", &token);" wasn't cutting it either.
>
> Thanks.

import std.stdio;
import std.array;
void main(){
    auto tokens = stdin.readln('@').split;
    writeln(tokens);
}

["1", "2", "3", "#", "$", "4", "3", "*", "!", "#", "20", "3", "/", "#", "$", "#", "62", "#", "$", "2", "3", "8", "*", "+", "#", "4", "48", "4", "2", "+", "/", "#", "SUM", "#", "$", "1", "2", "3", "4", "5", "#", "R", "#", "@"]

On Tuesday, 21 April 2015 at 03:44:16 UTC, weaselcat wrote:
> snip

Wow, that's a damn good solution... I didn't know that readln() could take an argument that it stops at once it finds.

Now the thing is, this program is supposed to be a reverse Polish notation calculator. A human using this program would probably be confused as to why nothing happens when they hit enter after a line -- it only really works in the context of copying and pasting the whole input in. Still a really neat solution to know anyhow. Thanks!