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I have this code in c language, it does check if a number written in a certain numeric base, decimal, octal, ..etc is correct, means that it is using characters which belongs to this certain base, for example, an octal number should only use characters [0, 1, 2, 3, 4, 5, 6, 7], it checks all the bases between 2 and 36.

The problem is that when I try to substring "base" characters from the total characters it give me a warning saying that ISO C90 forbids variable length array 'base_symbols'

int checkNumBase(char *num, int base){

        char all_symbols[36] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        char base_symbols[base];

        int i;
        unsigned int k;    

        for(i = 0; i<base; i++){
            base_symbols[i] = all_symbols[i];
        }


        for(k = 0; k<strlen(num); k++){        
            if(strchr(base_symbols, num[k]) == NULL){

                return 0;
            }
        }
        return 1;
    }
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  • 1
    Just replace char base_symbols[base]; with char base_symbols[36];, if you don't want to use VLAs. Commented Apr 2, 2019 at 10:01
  • 2
    I honestly don't see why you need base_symbols at all. The same indices are valid (and mean the same) in all_symbols. Commented Apr 2, 2019 at 10:02
  • I guess I need it, let's say the base is octal then I have substring 8 characters and check if my number belongs to them, please correct me if I am wrong Commented Apr 2, 2019 at 10:04
  • 1
    -ansi is the same as -std=c90 Commented Apr 2, 2019 at 10:17
  • 1
    -ansi means "give me 30 years old crap mode". You should be using gcc -std=c17 -pedantic-errors -Wall -Wextra. Commented Apr 2, 2019 at 10:34

4 Answers 4

Reset to default
4

One simple solution would be to truncate the string

char all_symbols[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";   // length corrected
if(base > 36)
    return 0;
all_symbols[base] = 0;
//.. as before
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4 Comments

By far the best solution. Some nitpick: It seems all_symbols shall be [37]
@4386427 good spot, I just copied that from the question. Edited.
could you please elaborate more I don't really know how this can work :P
Well, what you did was to copy base characters from all_symbols[] to base_symbols[], and this achieves the same thing: the required string of valid characters. But note: you haven't allowed room for the NUL string terminator with arrays of length [36] and [base], and you did not put a terminator at the end of base_symbols[].
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ISO C90 forbids variable length array 'base_symbols'

There's nothing wrong with the code, you get error this because you are using an old, obsolete compiler. You need to get a modern one such as gcc.

Please note that older versions of gcc did support newer versions of the language if you compiled correctly: gcc -std=c11 or gcc -std=c99, but used "gnu90" as default, which is C90 + non-standard extensions. Newer versions 5.0 or later default to "gnu11".

For example, -ansi means "give me 30 years old crap mode" aka C90. Unless you really need C90 for backwards-compatibility reasons, you should be using gcc -std=c17 -pedantic-errors -Wall -Wextra.

What is the difference between C, C99, ANSI C and GNU C?

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As of gcc 10.2.0, "gnu17" (the GNU dialect of ISO C17) is the default.
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use char *index;
then index = strchr(all_symbols, toupper ( num[k])); to see if the character is in the set
if index is in the set it will have a larger address. subtract the smaller address from the larger address to get a positive result
then if ( index && index - all_symbols < base) then num[k] is valid for that base.
toupper() is in ctype.h

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but why index - all_symbols and not the contrary?
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The solution by @WeatherVane (i.e. https://stackoverflow.com/a/55472654/4386427) is a very good solution for the code posted by OP.

The solution below shows an alternative approach that doesn't use string functions.

// Calculate the minimum base that allows use of char c
int requiredBase(char c)
{
  if (c >= '0' && c <= '9') return c - '0' + 1;  // '0' requires base 1, '1' requires base 2, ...
  if (c >= 'A' && c <= 'Z') return c - 'A' + 11; // 'A' requires base 11, 'B'requires base 12, ...
  return INT_MAX;
}

int checkNumBase(char *num, int base){
  while (*num)
  {
    if (requiredBase(*num) > base) return 0;
    ++num;
  }
  return 1;
}
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