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.

include c:\cxpl\codes;
real Val;
 
proc R2CF(N1, N2, Lev);         \Output continued fraction for N1/N2
int  N1, N2, Lev;
int  Quot, Rem;
[if Lev=0 then Val:= 0.0;
Quot:= N1/N2;
Rem:= rem(0);
IntOut(0, Quot);
if Rem then [ChOut(0, if Lev then ^, else ^;);  R2CF(N2, Rem, Lev+1)];
Val:= Val + float(Quot);        \generate value from continued fraction
if Lev then Val:= 1.0/Val;
];
 
int I, Data;
[Data:= [1,2, 3,1, 23,8, 13,11, 22,7, 0];
Format(0, 15);
I:= 0;
while Data(I) do
   [IntOut(0, Data(I));  ChOut(0, ^/);  IntOut(0, Data(I+1));  ChOut(0, 9\tab\);
   ChOut(0, ^[);  R2CF(Data(I), Data(I+1), 0);  ChOut(0, ^]);  ChOut(0, 9\tab\);
   RlOut(0, Val);  CrLf(0);
   I:= I+2];
]