11

I need to get an array size from user input. It seemed natural to me to store the input as size_t, however looking for an appropriate strto...() function I couldn't find any. I just used strtoull(), since unsigned long long is guaranteed to be at least 64 bits and I'm using C99 anyway. But I was wondering what would be the best way to get size_t from a string - say, in ANSI C.

Edit:

To clarify, I don't want the string length! The user will input the size of a large buffer in the form of a string, for instance "109302393029". I need to get that number and store as size_t. Of course I could use strtol() or even atoi(), but it seems like a clumsy hack (size_t may hold larger values than int, for instance, and I want the user to be able to input any value in the addressable space).

Sourav Ghosh
  • 133,132
  • 16
  • 183
  • 261
Alex
  • 1,416
  • 4
  • 16
  • 42
  • 1
    `strlen` will not work for a pointer to an array. – Tim Biegeleisen Sep 01 '15 at 07:08
  • 1
    You could use `unsigned long long` as the variable type. Seems unlikely that `size_t` would be larger than that. – M.M Sep 01 '15 at 07:12
  • 1
    possible duplicate of [printf format specifiers for uint32\_t and size\_t](http://stackoverflow.com/questions/3168275/printf-format-specifiers-for-uint32-t-and-size-t) – perreal Sep 01 '15 at 07:27

3 Answers3

16

In case you have a string input containing the value for a size_t and you want to get the value, you can use sscanf() with %zu format specifier to read the value and store it into corresponding size_t variable.

Note: do not forget to check the success of sscanf() and family.

Pseudo-code:

size_t len = 0;
if (1 == sscanf(input, "%zu", &len))
printf("len is %zu\n", len);

However, FWIW, this won't handle the overflow case. In case of the input length being arbitaryly large which your program should be able to handle, you may want to make use of strtoumax() and check for overflow, finally casting the returned value to size_t. Please see Mr. Blue Moon's answer related to this.

However, if you don't mind another approach, instead of taking the input as sting and converting it to size_t, you can directly take the input as size_t, like

size_t arr_size = 0;
scanf("%zu", &arr_size);
Community
  • 1
  • 1
Sourav Ghosh
  • 133,132
  • 16
  • 183
  • 261
  • 4
    He could also take the input as string and then use `sscanf` – M.M Sep 01 '15 at 07:13
  • The problem with this approach is that it's not safe against integer overflow, considering OP wants to handle arbitrarily large inputs. – P.P Sep 01 '15 at 07:27
  • @BlueMoon Sir, I think you meant _unsafe_, and that's very true. Range checking is needed in case of over/underflow. I'll add a paragraph about `strtoumax()` in my answer. Thanks for the suggestion. :) – Sourav Ghosh Sep 01 '15 at 07:31
  • Yes. I have already added an answer with `strtoumax` approach. – P.P Sep 01 '15 at 07:32
5

You can use strtoumax() to perform the conversion:

   #include <inttypes.h>

   intmax_t strtoimax(const char *nptr, char **endptr, int base);
   uintmax_t strtoumax(const char *nptr, char **endptr, int base);

and cast the result to size_t. This is a better approach since it helps detect overflow when arbitrarily large input is given.

The scanf() family functions can't detect integer overflow, which results in undefined behaviour.

P.P
  • 117,907
  • 20
  • 175
  • 238
2

Just get the input as unsigned int and cast it to size_t, or just:

size_t length;
scanf("%zu", &length);
perreal
  • 94,503
  • 21
  • 155
  • 181