High Level Function Confusion with Lambda/LINQ expressions in C# -

unsure how describe question title might wrong.

am reading on code examples , confused on following return function:

func<func<int , bool>, func<int , int>, func<int , int>> loop = null ; loop = (c , f ) => n => c(n) ? loop(c , f ) ( f (n)): n; func<int , int> w = loop(n => n < 10 , n => n + 2);  var r = w(2);  var s = w(3); console . writeline ("{0} {1}" , r , s ); 

i understand function returning loop when c(n) evaluates true, don't understand how loop(c,f) (f(n)) evaluates - both being passed loop? have tried running dumps in linqpad , don't how bit running.

any appreciated, understand dumb question!

one way try understand start small: have basic loop 1-10 +1 increment.

 func<int,int> basicloop = null;  basicloop = n => n < 10 ? basicloop(n+1) : n; 

that's pretty simple - basicloop function based on parameter n returns n (for n >= 10) or calls incremented parameter. basicloop(8) computed as:

• basicloop(8) 8 < 10, calls basicloop(8+1) result
• basicloop(9) 9 < 10, calls basicloop(9+1) result
• basicloop(10) 10 == 10, returns n 10.
• basicloop(9) got result 10 (from basicloop(10)) , returns it
• basicloop(8) got result 10 (from basicloop(9))and returns it

now want pass condition parameter loop. means our "loop" func need pass condition around on every iteration:

type of condition (similar n<10) - func<int, bool>. have takes func<int,bool> argument , returns same value our original basicloop. hence func of 1 argument , 1 result:

func<func<int, bool>, func<int,int>> condloop = null; 

condloop function of 1 argument - when defining take argument: condloop = (condition) => ....

we need replace condition in our original basicloop: n => n < 10 ? ... becomes n => condition(n) ? ....

the last part replacing basicloop(n+1) - have condloop function returns equivalent of basicloop when pass condition it. fortunately our condition not change between iterations , have - condloop(condition) equivalent of basicloop. getting together:

condloop = (condition) =>    n => condition(n) ?        condloop(condition) (n + 1) :       n;  

tracing through calls condloop(x => x < 5)(4)

• condloop(x => x < 5)(4) - condition x => x < 5, n = 4 when condition(4) called x = 4, 4 < 5 true - calling condloop same condition , increased n - condloop(x => x < 5)(4 + 1) result
• condloop(x => x < 5)(5) - condition x => x < 5, n = 5 when condition(5) called x = 5, 5 < 5 false - returning n 5
• back condloop(x => x < 5)(4) - returning 5 result of condloop(x => x < 5)(5)

with similar logic adding function increments value - on every iteration need pass condition , increment function (c , f in original post).