Remove spaces from a (copy of) a proto when used. The logic that uses…

… prototypes assumes spaces were already gone, which may not be true if they were added via XS / set_prototype.

inline.h

@@ -32,6 +32,34 @@ S_CvDEPTHp(const CV * const sv)
     return &((XPVCV*)SvANY(sv))->xcv_depth;
 }
 
+/*
+ CvPROTO returns the prototype as stored, which is not necessarily what
+ the interpreter should be using. Specifically, the interpreter assumes
+ that spaces have been stripped, which has been the case if the prototype
+ was added by toke.c, but is generally not the case if it was added elsewhere.
+ Since we can't enforce the spacelessness at assignment time, this routine
+ provides a temporary copy at parse time with spaces removed.
+ I<orig> is the start of the original buffer, I<len> is the length of the
+ prototype and will be updated when this returns.
+ */
+
+PERL_STATIC_INLINE char *
+S_strip_spaces(pTHX_ const char * orig, STRLEN * const len)
+{
+    SV * tmpsv;
+    char * tmps;
+    tmpsv = newSVpvn_flags(orig, *len, SVs_TEMP);
+    tmps = SvPVX(tmpsv);
+    while ((*len)--) {
+	if (!isSPACE(*orig))
+	    *tmps++ = *orig;
+	orig++;
+    }
+    *tmps = '\0';
+    *len = tmps - SvPVX(tmpsv);
+		return SvPVX(tmpsv);
+}
+
 /* ----------------------------- regexp.h ----------------------------- */
 
 PERL_STATIC_INLINE struct regexp *

op.c

@@ -10078,6 +10078,7 @@ Perl_ck_entersub_args_proto(pTHX_ OP *entersubop, GV *namegv, SV *protosv)
     if (SvTYPE(protosv) == SVt_PVCV)
 	 proto = CvPROTO(protosv), proto_len = CvPROTOLEN(protosv);
     else proto = SvPV(protosv, proto_len);
+    proto = S_strip_spaces(aTHX_ proto, &proto_len);
     proto_end = proto + proto_len;
     aop = cUNOPx(entersubop)->op_first;
     if (!aop->op_sibling)

toke.c

@@ -7281,6 +7281,7 @@ Perl_yylex(pTHX)
 			STRLEN protolen = CvPROTOLEN(cv);
 			const char *proto = CvPROTO(cv);
 			bool optional;
+			proto = S_strip_spaces(aTHX_ proto, &protolen);
 			if (!protolen)
 			    TERM(FUNC0SUB);
 			if ((optional = *proto == ';'))