Problem: Vim9: cannot use 0 or 1 where a bool is expected.
Solution: Allow using 0 and 1 for a bool type. (closes #6903)
/* vim9type.c */
+type_T *alloc_type(garray_T *type_gap);
void clear_type_list(garray_T *gap);
type_T *get_list_type(type_T *member_type, garray_T *type_gap);
type_T *get_dict_type(type_T *member_type, garray_T *type_gap);
#define TTFLAG_VARARGS 1 // func args ends with "..."
#define TTFLAG_OPTARG 2 // func arg type with "?"
+#define TTFLAG_BOOL_OK 4 // can be converted to bool
/*
* Structure to hold an internal variable without a name.
let bool2: bool = false
assert_equal(v:false, bool2)
+ let bool3: bool = 0
+ assert_equal(0, bool3)
+ let bool4: bool = 1
+ assert_equal(1, bool4)
+
CheckDefFailure(['let x:string'], 'E1069:')
CheckDefFailure(['let x:string = "x"'], 'E1069:')
CheckDefFailure(['let a:string = "x"'], 'E1069:')
static int included_patches[] =
{ /* Add new patch number below this line */
+/**/
+ 1641,
/**/
1640,
/**/
generate_PUSHNR(cctx_T *cctx, varnumber_T number)
{
isn_T *isn;
+ garray_T *stack = &cctx->ctx_type_stack;
RETURN_OK_IF_SKIP(cctx);
if ((isn = generate_instr_type(cctx, ISN_PUSHNR, &t_number)) == NULL)
return FAIL;
isn->isn_arg.number = number;
+ if (number == 0 || number == 1)
+ {
+ type_T *type = alloc_type(cctx->ctx_type_list);
+
+ // A 0 or 1 number can also be used as a bool.
+ if (type != NULL)
+ {
+ type->tt_type = VAR_NUMBER;
+ type->tt_flags = TTFLAG_BOOL_OK;
+ ((type_T **)stack->ga_data)[stack->ga_len - 1] = type;
+ }
+ }
return OK;
}
* Allocate memory for a type_T and add the pointer to type_gap, so that it can
* be freed later.
*/
- static type_T *
+ type_T *
alloc_type(garray_T *type_gap)
{
type_T *type;
{
if (expected->tt_type != actual->tt_type)
{
+ if (expected->tt_type == VAR_BOOL && actual->tt_type == VAR_NUMBER
+ && (actual->tt_flags & TTFLAG_BOOL_OK))
+ // Using number 0 or 1 for bool is OK.
+ return OK;
if (give_msg)
arg_type_mismatch(expected, actual, argidx);
return FAIL;