Fix operator parsing

Author: tautschnig

regression/cpp/Function_Overloading3/main.cpp

@@ -0,0 +1,21 @@
+#include <cassert>
+
+class t1{
+    public:
+        t1 (int n) :value(n) {}
+    public:
+        int value;
+        int operator[](int n) { return n*value; }
+};
+
+int operator+(t1 left, int right) {return left.value+right;}
+
+int main()
+{
+    t1 t(10);
+    int t_1 = t + 5;
+    int t_2 = t[5];
+    assert(t_1 == 15);
+    assert(t_2 == 50);
+    return 0;
+}

regression/cpp/Function_Overloading3/test.desc

@@ -0,0 +1,8 @@
+CORE
+main.cpp
+
+^EXIT=0$
+^SIGNAL=0$
+--
+^warning: ignoring
+^CONVERSION ERROR$

regression/cpp/Unary_Function_Overload3/test.desc

@@ -1,4 +1,4 @@
-KNOWNBUG
+CORE
 main.cpp
 
 ^EXIT=0$

src/cpp/parse.cpp

@@ -3578,7 +3578,7 @@ bool Parser::rOperatorName(irept &name)
   case '<':
   case '>':
   case ',':
-    operator_id=irep_idt(std::string(static_cast<char>(t), 1));
+    operator_id=irep_idt(std::string(1, static_cast<char>(t)));
     break;
 
   case TOK_MULTASSIGN: operator_id="*="; break;