Detecting if a string is a double or long in C

Question

I'm dealing with an char[] containing text which should represent a double number value OR a long number.
I'm in a need to write a function that detects which of the above data-types is represented (if any).

I thought about using strtol() and check if it fails to parse the entire string, and if it fails, using strtod().

I would be happy to see whether there is a better option to do so. Thanks.

Answer 1

I thought about using strtol() and check if it fails to parse the entire string, and if it fails, using strtod().

I think that's a good idea and I don't think there is a better one. Implementing your own parsing routine is generally a bad idea.

I would trim the string from trailing whitespace before calling strtol to avoid false negatives.

Answer 2

strtol() and strtod() is the right approach. Be sure to use errno to detect integer overflow. 2 stand-alone functions follow:

int  Is_long(const char *src, long *dest) {
  char *endptr;
  // Clear, so it may be tested after strtol().
  errno = 0;  
  // Using 0 here allows 0x1234, octal 0123 and decimal 1234. 
  long num = strtol(src, &endptr, 0);
  // If +/- overflow, "" or has trailing text ...
  if (errno || endptr == src || *endptr != '\0') {
    return 0;
  }
  if (dest) *dest = num;
  return 1;
}

int  Is_double(const char *src, double *dest) {
  char *endptr;
  // In this case, detecting over/undeflow IMO is not a concern, so ignore it.
  double num = strtod(src, &endptr);
  // If "" or has trailing text ...
  if (endptr == src || *endptr != '\0') {
    return 0;
  }
  if (dest) *dest = num;
  return 1;
}

@Klas Lindbäck does bring up the good point of what to do about trailing white-space. This answer assumes it is not valid.

Answer 3

 You can use the following code to detect that.

 char* isDouble = strchr(string, '.');
 if (isDouble) { 
   // is Double here
 }else {
   // is long here
 }