On Sat, 6 Mar 2021 07:25:18 +0530
Sameeruddin Shaik <sameeruddin.shaik8@xxxxxxxxx> wrote:
> i have one doubt.
> >Note, @filters should be of type: const char * const * filters, as not
> >only is filters pointing to constant strings, the array itself will not be
> >modified.
>
> what If the user wants to capture the filters at run time like below ?
> let's say
>
> filters = malloc(sizeof(char *));
> if (!filters)
> return 1;
> printf("please enter the input filters count\n");
> scanf("%d", &fil_count);
> while(i < fil_count) {
> scanf("%s", buf);
> slen = strlen(buf);
> if (!slen)
> return 1;
> filters[i] = calloc(1, slen);
> strncpy(filters[i++], buf, slen);
> }
> at that time, this declaration will be problematic right?, because we
> are trying to modify
> the read-only memory. Are we expecting the user to supply filters at
> compile time like below?
> const char * const *filters = {"kvm_pmu_reset", "kvm_pmu_init",
> "dir_item_err", NULL};
OK, my apologies, I see the issue that you are having, and you are
correct.
Because newer compiles will warn if you pass "char **" to a
"const char * const *" parameter. Because it assumes that the two types
are different, even when they shouldn't be. I'm not sure why the
compiler wont let you pass in a char ** to a const char * const *, but
it does indeed make them different.
Even though there's ways to always pass strings via this logic, (you
can create a "const char **" array and assign it to the dynamic one,
and pass that in just fine). I looked at other prototypes, and see that
the common method is.
const char **
A couple do the "const char * const *" but they look to be special
cases.
So yes, let's go with "const char **" as we want to show that the
strings will not be modified, and have "const char ***" for errs.
Thanks!
-- Steve
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