Re: ${(z)} parsing of multiple array assignments

[Peter Stephenson <p.w.stephenson@xxxxxxxxxxxx>]

On Mon, 2019-12-23 at 17:31 +0000, Daniel Shahaf wrote:
> In the following two cases, why are the assignments to $b
> parsed differently to the assignments to $a?
> 
>     % pz() { print -rl -- ${(qqqq)${(z)1}} }
> 
>     % pz 'a=() b=()' 
>     $'a=('
>     $')'
>     $'b='
>     $'()'
> 
>     % pz 'a=(foo) b=(bar)' 
>     $'a=('
>     $'foo'
>     $')'
>     $'b=(bar)'

Unless I'm missing some trick, bufferwords() is the function where we
need to update any parser state --- ctxtlex() is too low level for that,
it just handles the next token given the current state.  So it's
probably something like this.  Will need a new test adding.

Presumably the commenting out of ENVSTRING in ctxtlex() means we stay in
incmdpos in that case --- as this is just a single word assignment that
probably handles that case OK.

pws

diff --git a/Src/hist.c b/Src/hist.c
index 74116e82f..37cba4579 100644
--- a/Src/hist.c
+++ b/Src/hist.c
@@ -3321,6 +3321,7 @@ bufferwords(LinkList list, char *buf, int *index, int flags)
     int owb = wb, owe = we, oadx = addedx, onc = nocomments;
     int ona = noaliases, ocs = zlemetacs, oll = zlemetall;
     int forloop = 0, rcquotes = opts[RCQUOTES];
+    int envarray = 0;
     char *p, *addedspaceptr;
 
     if (!list)
@@ -3404,6 +3405,14 @@ bufferwords(LinkList list, char *buf, int *index, int flags)
 	ctxtlex();
 	if (tok == ENDINPUT || tok == LEXERR)
 	    break;
+	/*
+	 * After an array assignment, return to the initial
+	 * start-of-command state.  There could be a second ENVARRAY.
+	 */
+	if (tok == OUTPAR && envarray) {
+	    incmdpos = 1;
+	    envarray = 0;
+	}
 	if (tok == FOR) {
 	    /*
 	     * The way for (( expr1 ; expr2; expr3 )) is parsed is:
@@ -3441,6 +3450,7 @@ bufferwords(LinkList list, char *buf, int *index, int flags)
 	    switch (tok) {
 	    case ENVARRAY:
 		p = dyncat(tokstr, "=(");
+		envarray = 1;
 		break;
 
 	    case DINPAR: