Differences

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--- language:controlflow [2020/06/03 18:21] – [Selections] rajit
+++ language:controlflow [2022/07/17 12:07] (current) – [Loops] rajit
@@ Line 11: / Line 11: @@
 An example of the loop construct in ACT is shown below:
-<code>
+<code act>
 ( i : 10 : bool x[i..i]; )
 </code>
@@ Line 20: / Line 20: @@
 range ''0..9''. The body of the loop is the statement ''bool x[i..i];''. The effect of this statement is the same as
-<code>
+<code act>
 bool x[0..0];
 bool x[1..1];
@@ Line 31: / Line 31: @@
 is shown below:
-<code>
+<code act>
 register r[1..8];
 (i : 1..8 : r[i](in[i],out[i],control); )
@@ Line 48: / Line 48: @@
 command language.
-<code>
+<code act>
 pint i;
 i=0;
@@ Line 65: / Line 65: @@
 In production rule bodies, the loop
-<code>
+<code act>
 (&i:3: x[i])
 </code>
 expands to
-<code>
+<code act>
 x[0] & x[1] & x[2]
 </code>
@@ Line 75: / Line 75: @@
 syntax). The ''&'' symbol separates the body of the loop that is
 instantiated for different values of ''i''.
+The syntactic replication construct is written as follows:
+<code act>
+(sym id : range : body )
+</code>
+The ''sym'' (symbol) might be empty. ''id'' is a variable that can be used in ''body'', and takes the range specified by ''range''. ''range'' can be either an integer-valued expression or ''start .. end'' to indicate a start and end index. The result of the replication is
+<code act>
+ body(id set to lo) sym body(id set to lo+1) sym ... sym body(id set to hi)
+</code>
+where ''lo'' is the starting index of the range, and ''hi'' is the ending index, and ''body(id set to x)'' means ''body'' with the value of ''x'' substituted by the constant ''id''.
@@ Line 82: / Line 92: @@
 syntax of a selection statement is:
-<code>
+<code act>
 [ boolean_expression -> body
 [] boolean_expression -> body
@@ Line 95: / Line 105: @@
 something special for register 0 as follows:
-<code>
+<code act>
 (i : 32 :
     [ i = 0 -> r0(in[i],out[i],control);
@@ Line 116: / Line 126: @@
 definition can be used to create a tree structure.
-<code>
+<code act>
 template<pint N>
 defproc tree (bool a[N])
 {
-  [ N = 1 -> leaf l(a[0])
+  [ N = 1 -> leaf l(a[0]);
  [] N > 1 -> tree<N/2> t0(a[0..N/2-1]);
-             tree<N-N/2> t1(a[N/2..N-1])
+             tree<N-N/2> t1(a[N/2..N-1]);
   ]
 }