Ævar Arnfjörð Bjarmason <avarab@xxxxxxxxx> writes:
> That I just don't understand, i.e. how the variable is defined in this
> 6/5 pertains to how it gets used /inside/ the function in this case, the
> caller code doesn't have to care, but perhaps I'm missing something...
-static struct grep_pat *create_grep_pat(const char *pat, size_t patlen,
+static struct grep_pat *create_grep_pat(const char *const pat, size_t patlen,
const char *origin, int no,
enum grep_pat_token t,
enum grep_header_field field)
Because the function signature is visible to the caller of this
function, it sees that the callee, as part of its implementation
detail, keeps "pat" pointer pointing at the same location. That is
not something the caller needs to care, even though the caller may
deeply care that the memory locations "pat" points at will not be
changed via the pointer by calling this function.
On the other hand
@@ -1438,7 +1439,7 @@ static int look_ahead(struct grep_opt *opt,
const char **bol_p)
{
unsigned lno = *lno_p;
- const char *bol = *bol_p;
+ const char * const bol = *bol_p;
struct grep_pat *p;
const char *sp, *last_bol;
regoff_t earliest = -1;
This one is totally invisible to the caller of the function, and is
a good documentation for those who read the implementation of it.
Thanks.
