Continued fraction arithmetic
The purpose of this task is to write a function 





r
2
c
f


(

i
n
t



{\displaystyle {\mathit {r2cf}}(\mathrm {int} }

 




N

1


,

i
n
t



{\displaystyle N_{1},\mathrm {int} }

 




N

2


)


{\displaystyle N_{2})}

, or 





r
2
c
f


(

F
r
a
c
t
i
o
n



{\displaystyle {\mathit {r2cf}}(\mathrm {Fraction} }

 



N
)


{\displaystyle N)}

, which will output a continued fraction assuming:






N

1




{\displaystyle N_{1}}

 is the numerator





N

2




{\displaystyle N_{2}}

 is the denominator
The function should output its results one digit at a time each time it is called, in a manner sometimes described as lazy evaluation.

To achieve this it must determine: the integer part; and remainder part, of 




N

1




{\displaystyle N_{1}}

 divided by 




N

2




{\displaystyle N_{2}}

. It then sets 




N

1




{\displaystyle N_{1}}

 to 




N

2




{\displaystyle N_{2}}

 and 




N

2




{\displaystyle N_{2}}

 to the determined remainder part. It then outputs the determined integer part. It does this until 




a
b
s

(

N

2


)


{\displaystyle \mathrm {abs} (N_{2})}

 is zero.

Demonstrate the function by outputing the continued fraction for:

 1/2
 3
 23/8
 13/11
 22/7
 -151/77






2




{\displaystyle {\sqrt {2}}}

 should approach 



[
1
;
2
,
2
,
2
,
2
,
…
]


{\displaystyle [1;2,2,2,2,\ldots ]}

 try ever closer rational approximations until boredom gets the better of you:

 14142,10000
 141421,100000
 1414214,1000000
 14142136,10000000
Try :

 31,10
 314,100
 3142,1000
 31428,10000
 314285,100000
 3142857,1000000
 31428571,10000000
 314285714,100000000
Observe how this rational number behaves differently to 





2




{\displaystyle {\sqrt {2}}}

 and convince yourself that, in the same way as 



3.7


{\displaystyle 3.7}

 may be represented as 



3.70


{\displaystyle 3.70}

 when an extra decimal place is required, 



[
3
;
7
]


{\displaystyle [3;7]}

 may be represented as 



[
3
;
7
,
∞
]


{\displaystyle [3;7,\infty ]}

 when an extra term is required.

import std.concurrency;
import std.stdio;
 
struct Pair {
    int first, second;
}
 
auto r2cf(Pair frac) {
    return new Generator!int({
        auto num = frac.first;
        auto den = frac.second;
        while (den != 0) {
            auto div = num / den;
            auto rem = num % den;
            num = den;
            den = rem;
            div.yield();
        }
    });
}
 
void iterate(Generator!int seq) {
    foreach(i; seq) {
        write(i, " ");
    }
    writeln();
}
 
void main() {
    auto fracs = [
        Pair(   1,  2),
        Pair(   3,  1),
        Pair(  23,  8),
        Pair(  13, 11),
        Pair(  22,  7),
        Pair(-151, 77),
    ];
    foreach(frac; fracs) {
        writef("%4d / %-2d = ", frac.first, frac.second);
        frac.r2cf.iterate;
    }
    writeln;
 
    auto root2 = [
        Pair(    14_142,     10_000),
        Pair(   141_421,    100_000),
        Pair( 1_414_214,  1_000_000),
        Pair(14_142_136, 10_000_000),
    ];
    writeln("Sqrt(2) ->");
    foreach(frac; root2) {
        writef("%8d / %-8d = ", frac.first, frac.second);
        frac.r2cf.iterate;
    }
    writeln;
 
    auto pi = [
        Pair(         31,          10),
        Pair(        314,         100),
        Pair(      3_142,       1_000),
        Pair(     31_428,      10_000),
        Pair(    314_285,     100_000),
        Pair(  3_142_857,   1_000_000),
        Pair( 31_428_571,  10_000_000),
        Pair(314_285_714, 100_000_000),
    ];
    writeln("Pi ->");
    foreach(frac; pi) {
        writef("%9d / %-9d = ", frac.first, frac.second);
        frac.r2cf.iterate;
    }
}