How do I tell if the c function atoi failed or if it was a string of zeros?

Question

When using the function atoi (or strtol or similar functions for that matter), how can you tell if the integer conversion failed or if the C-string that was being converted was a 0?

For what I'm doing, 0 is an acceptable value and the C-string being converted may contain any number of 0s. It may also have leading whitespace.

Answer 1

The proper function (as long as you are insisting on using C-style functions) is strtol and the conversion code might look as follows

const char *number = "10"; /* for example */  char *end; long value = strtol(number, &end, 10);  if (end == number || *end != '\0' || errno == ERANGE)   /* ERROR, abort */;  /* Success */ /* Add whatever range checks you want to have on the value of `value` */ 

Some remarks:

strtol allows (meaning: quietly skips) whitespace in front of the actual number. If you what to treat such leading whitespace as an error, you have to check for it yourself.

The check for *end != '\0' makes sure that there's nothing after the digits. If you want to permit other characters after the actual number (whitespace?), this check has to be modified accordingly.

P.S. I added the end == number check later to catch empty input sequences. "All whitespace" and "no number at all" inputs would have been caught by *end != '\0' check alone. It might make sense to catch empty input in advance though. In that case end == number check will/might become unnecessary.

Answer 2

Since this is tagged c++:

template< typename T > inline T convert(const std::string& str) {     std::istringstream iss(str);     T obj;      iss >> std::ws >> obj >> std::ws;      if(!iss.eof())         throw "dammit!";      return obj;  }