Learning Forum

[fdtsaid]

In Scade Language Reference 8.5 Operator Application and Higher-Order Patterns, it illustrated that the usage of foldw:

 

idx, acc = (foldw op << size >> if cond0)(acc0, A1,...,An)

 

is equivalent to particular form of mapfoldw:

 

idx, _, acc = (mapfoldw 1 op << size >> if cond0)(acc0, A1, ..., An)

 

Should it be the following form instead?

idx, _, acc = (mapfoldw 1 op << size >> if cond0 default ())(acc0, A1, ..., An)

 

Because the syntax definition of mapfoldw is 

operator ::= (iterator_mw operator << size >> if expr default expr)

iterator_mw ::= mapfoldw [[ INTEGER ]]

[Benjamin Descorps]

Hello,

You're correct. The right syntax is

idx, _, acc = (mapfoldw 1 op <> if cond0 default ()) (acc0, A1,…,An)

Let's have the following example:

const n : uint8 = 10;

function Op (c,d: bool) returns (cond_o, s: bool)

let 

s = c or d;

cond_o = not s;

tel 

 

function disjunction(B: bool^n) returns (exist: bool; i: bool)

i, exist = (foldw Op <> if true) (false, B);

 

The foldw could be replaced by the mapfoldw in the following way:

i, _, exist = (mapfoldw 1 Op <> if true default ())(false, B);

 

Regards,

 

Benjamin