Focusing only on strtoul() here, the correct error checking in most cases is:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | unsigned long mystrtoul(char *str, bool *success) { unsigned long val = 0; char *eptr = NULL; errno = 0; *success = true; val = strtoul(str, &eptr, 0); if (eptr == str || *eptr != '\0' || (val == ULONG_MAX && errno == ERANGE) || (val == 0 && errno == EINVAL)) *success = false; return val; } |
That is:
- IF your end pointer points to the beginning of your string, you've got a problem
- IF your end pointer doesn't point to the end of your string, you've got a problem
- IF your output is at the max end of the range (a potentially valid value) AND you get errno, you've got a problem
- IF your output is 0 (a potentially valid value) AND you get errno, you've got a problem
- ELSE congratulations, you've converted a string to an integer
If you originally wrote this code for a 32-bit system, with the intention of converting and storing a 32-bit number, then what you wrote is correct. If you then recompile this correct code on a 64-bit system, it becomes incorrect.
That's right with no modifications (and no compiler warnings), your carefully reviewed integer conversion function goes from right to (dangerously) wrong.
For example give as input "4294967296" and assign to an unsigned int:
1 2 3 4 5 6 7 8 9 10 11 | int main(void) { bool success = false; unsigned int val = 0; val = mystrtoul("4294967296", &success); printf("%s %u\n", success?"success":"failure", val); return 0; } |
Result:
- On 32-bit Linux: failure 4294967296 (ERANGE)
- On 64-bit Linux: success 0
- malloc(4294967296) allocate 0 bytes and a buffer overflow is created
- uid_t 4294967296 become 0 and a user is root
This isn't strictly a strtoul() problem. This is a LP64 data model problem. Any assumption about the 32-bit size of a long needs to be checked when converting from 32-bit to 64-bit runtime. This is just a particularly tricky example of the trade offs made by LP64.
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