fix result propogation in constexpr operator chains

## compile-time chained operator evaluation mechanism

Constants are stored in `value` object. The object has two members: `constval` and `boolresult`. The `constval` member stores the actual value which the object represents: `15`  and `10` for example in `15 < 20`.  All operators are in a chain of operators. On most occasions, there is just one operator in the chain but it is still a valid singleton chain. The `boolresult` member stores the result of the previous operations which the constant was involved in if any. For example, in `15 < 20 < 30`, the `boolresult` member of object representing `20` stores the result of `15 < 20`. In more complex expressions such as, `15 < 20 < 30 < 40 < 50`, the `boolresult` member of object representing `40` stores the result of the `15 < 20 < 30 < 40`.

The `boolresult` of the first constant in a chain is set to `TRUE` by default.

**Details:**

The processing of chains begins at `plnge_rel` which maintains two `value` objects: `lval` and `lval2`. The `lval` is taken by `plnge_rel` as an argument of the first constant (stored in `constval` field) value in the operator chain. The result of the full chain is also passed through it by storing the result in the `constval` field (like `15 < 20` evaluates to the constant `1`; the result of `15 < 20` is `1` which is also a constant).

The `boolresult` of `lval` is set to `TRUE` as this is the first constant in the chain. The `plnge_rel` progresses through the chain one by one with `lval` and `lval2` always being the left operand and the right operand respectively. For each operator it calls `plnge2` which dumps instructions for loading constant and the right operand into the staging buffer. If the right operand turns out to be a constant, the staging buffer is flushed and `static cell calc(cell left,void (*oper)(),cell right,char *boolresult);` is called to evaluate the compile-time expression with the constant value of the left and right operand passed as `left` and `right` arguments along with the operator and `boolresult` of the left operand. `calc` evluates the expression and returns the right operand and sets the `boolresult` to the result of the operator chain. After the call returns, `constval` of `lval` (left operand) is set to the `constval` of `lval2` (right operand). As of now, the `lval` now stores the right operand and the result of the operator chain uptil now.

`plnge2` returns and control is transfered back to `plnge_rel`. `plnge_rel` copies the value stored in `lval2` into `lval`. This implies that the `boolresult` and `constval` fields of `lval2` are copied into `lval` as the previous right operand is now the left operand for the next operator in the chain. **Oops! BUG!** The `boolresult` which was stored in `lval` is lost and is replaced by `FALSE` which was stored in `lval2` which is the default for `value` objets (set by `clear_value`). The compiler should've ensured that the `boolresult` of `lval` was retained. `plnge2` is again called to process the right operand and evaluate the expression.

This keeps repeating until the entire chain has been evaluated. After the entire chain has been evaluated, `lval` has `constval` as the most recent right operand and `boolresult` has the result of the entire chain. The compiler sets `constval` of `lval` to `boolresult` of `lval` and its tag to `bool`. Note that this is done in the last step. The `lval` now is a constant representing the result of the entire chain of operators, i.e. `1` if the chain evaluated to `TRUE` or `0` otherwise.

**Relavant bugged code:**
```
static int plnge_rel(int *opstr,int opoff,int (*hier)(value *lval),value *lval)
{
  .
  .
  .
  lval->boolresult=TRUE;
  do {
    /* same check as in plnge(), but "chkbitwise" is always TRUE */
    if (count>0 && bitwise_opercount!=0)
      error(212);
    if (count>0) {
      relop_prefix();
      *lval=lval2;      /* copy right hand expression of the previous iteration */
    } /* if */
    opidx+=opoff;
    plnge2(op1[opidx],hier,lval,&lval2);
    if (count++>0)
      relop_suffix();
  } while (nextop(&opidx,opstr)); /* enddo */
  lval->constval=lval->boolresult;
  lval->tag=pc_addtag("bool");    /* force tag to be "bool" */
  return FALSE;         /* result of expression is not an lvalue */
}
```

```
static void plnge2(void (*oper)(void),
                   int (*hier)(value *lval),
                   value *lval1,value *lval2)
{
    .
    .
    .
    if (check_userop(oper,lval1->tag,lval2->tag,2,NULL,&lval1->tag)) {
      lval1->ident=iEXPRESSION;
      lval1->constval=0;
    } else if (lval1->ident==iCONSTEXPR && lval2->ident==iCONSTEXPR) {
      /* only constant expression if both constant */
      stgdel(index,cidx);       /* scratch generated code and calculate */
      check_tagmismatch(lval1->tag,lval2->tag,FALSE,-1);
      lval1->constval=calc(lval1->constval,oper,lval2->constval,&lval1->boolresult);
    } else {
    .
    .
    .
}
```

```
static cell calc(cell left,void (*oper)(),cell right,char *boolresult)
{
  .
  .
  .
  else if (oper==os_le)
    return *boolresult &= (char)(left <= right), right;
  else if (oper==os_ge)
    return *boolresult &= (char)(left >= right), right;
  else if (oper==os_lt)
    return *boolresult &= (char)(left < right), right;
  else if (oper==os_gt)
    return *boolresult &= (char)(left > right), right;
  .
  .
  .
}
```

As you can see, the `boolresult` is ANDed with the result of the current operation. If one of the operators in the chain earlier had evaluated to false, the `&` will force `boolresult` to remain `FALSE` irrespective of whether the current expression evaluates to true or not.

### What's the bug?

The compiler loses the `boolresult` of `lval` in `plnge_rel` and replaces it with the `boolresult` of `lval2` which is always initalized to `FALSE` by `clear_val`.

The fix is to ensure that `lval` retains the value.

This commits does this by copying the `boolresult` of `lval2` into `lval` before `*lval=lval2`. This ensures that the `boolresult` is retained in `lval`.

source/compiler/sc3.c

@@ -512,6 +512,7 @@ static int plnge_rel(int *opstr,int opoff,int (*hier)(value *lval),value *lval)
       error(212);
     if (count>0) {
       relop_prefix();
+      lval2.boolresult=lval->boolresult;
       *lval=lval2;      /* copy right hand expression of the previous iteration */
     } /* if */
     opidx+=opoff;