Reading in hexadecimal

Question

I am trying to read in hexadecimal from a text file, currently this is what I have

     void hexReader(char* file, node* head){
        FILE *fp;
        char str[MAXCHAR];

        size_t number = 0;

         fp = fopen(file, "r");
         if (fp == NULL){
         printf("Could not open file %s",filename);
        //return 1;
                                          }
           while (fgets(str, MAXCHAR, fp) != NULL) {
         //while ((number = getline(&line, &len, fp)) != -1)
             number = strtoull(str, NULL, 16);
             if (number > 0){
             printf("%size_t \n",number);
             printf("%x \n",number);

         }
       }
       fclose(fp);

  }

I have tried using unsigned long long for number but it fails on hex like FFFFFFFFFFFFFFF. I should be able to read in things like that, in the form of 0x....up to a 64 bit hex number.

Why is this failing? Size_t prints out nonesense and unsinged long long fails on too large of numbers.

`printf("%size_t \n",number);` seems off, `%s` means string. — Joachim Isaksson, Oct 09 '13 at 17:10
And [here](http://stackoverflow.com/questions/2125845/platform-independent-size-t-format-specifiers-in-c) is a question asking about what the format specifier for `size_t` is supposed to be. — Dennis Meng, Oct 09 '13 at 17:11
@JoachimIsaksson I can change that and I still get the wrong thing. — Paul the Pirate, Oct 09 '13 at 17:14
@PaulthePirate Looks like Carl's answer touches on why you might not be able to use `%zx`. — Dennis Meng, Oct 09 '13 at 17:22
@DennisMeng I am not sure how I can look that up in windows, whatever the newest codeblocks uses. — Paul the Pirate, Oct 09 '13 at 17:23
If you're using Windows, then don't worry about it (I'm pretty sure glibc applies to Linux and Mac only) — Dennis Meng, Oct 09 '13 at 17:24

chux - Reinstate Monica · Accepted Answer · 2013-10-09T17:38:53.270

1

Small re-write of inner loop.

1 No need for size_t.
2 Use errno to check for success of strtull().
3 Use correct unsigned long long specifier modifier "ll".
4 printf("%size_t \n",number) failed because the %s means to assume number is a string.
5 printf("%x \n",number) failed as sizeof(number) my be larger than sizeof(unsigned), which works with %x, was only using some of number.

while (fgets(str, MAXCHAR, fp) != NULL) {
  errno = 0;
  char *endptr;
  unsigned long long X = strtoull(str, &endptr, 16);
  // Use these tests as desired to catch overflow, no convert, unexpected extra.
  if (errno || (endptr == str) || (*endptr != '\n')) {
    printf("Error\n");
  }
  else {
    printf("%llx\n", X);  // print hex
    printf("%#llx\n", X);  // print hex with leading 0x
    printf("%#llu\n", X);  // print decimal
  }
}

edited Oct 09 '13 at 17:38

answered Oct 09 '13 at 17:21

chux - Reinstate Monica

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I don't think unsigned long long is going to cut it, it fails to work on larger values. – Paul the Pirate Oct 09 '13 at 17:22
@PaulthePirate, if `unsigned long long` won't work, your `strtoull` call is going to be a problem. It *by definition* is returning an `unsigned long long` to you. Is your `size_t` maybe not a 64-bit type? – Carl Norum Oct 09 '13 at 17:23
@CarlNorum Is there a size_t version of strt? Or is there any way I can store a line of text to compare directly? That might be easier. – Paul the Pirate Oct 09 '13 at 17:24
I've never heard of a system where `unsigned long long` wasn't as big or bigger than a `size_t`. Check yours! – Carl Norum Oct 09 '13 at 17:25
@CarlNorum It specifically fails on 0x999999999999999999999 – Paul the Pirate Oct 09 '13 at 17:27
1

`0x999999999999999999999` is much larger than 64-bits can hold. – Carl Norum Oct 09 '13 at 17:27
@CarlNorum How do I know that? – Paul the Pirate Oct 09 '13 at 17:27
1

@PaulthePirate I count 21 9s, which would need 11 bytes, or 88 bits. – Dennis Meng Oct 09 '13 at 17:28
1

One hexadecimal digit is equivalent to 4 bits (16 == 2^4). The largest 64-bit number is `0xFFFFFFFFFFFFFFFF`, which has 16 digits. Your number has 21 digits, so would need 84 bits to be stored. – Carl Norum Oct 09 '13 at 17:28
@Paul the Pirate By the C spec, `unsigned long long` is good for _at least_ 64 bits. – chux - Reinstate Monica Oct 09 '13 at 17:41
@Paul the Pirate There is a `strtoumax()` which returns type `uintmax_t`. The `printf()` specifier is `"%" PRIxMAX`. Your compiler likely does not support a larger integer type. This still may be too small for an 88-bit number. – chux - Reinstate Monica Oct 09 '13 at 17:44

score 1 · Answer 2 · answered Oct 09 '13 at 17:22

1

%size_t is nonsense - use %zx to print a hexadecimal size_t value. Your %x won't work either - if you don't have the z modifier because you're using a non-C99-aware compiler/library, make sure you match variable type and format string:

printf("%llx\n", (unsigned long long)number);

answered Oct 09 '13 at 17:22

Carl Norum

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Hah, thanks for clearing up why he might be having issues with `%zx`. – Dennis Meng Oct 09 '13 at 17:24

Reading in hexadecimal

2 Answers2