testing a string to see if a number is present and asigning that value to a variable while skipping all the non-numeric values?

Question

Hi there, given a string say " a 19 b c d 20", how do I test to see if at that particular position on the string there is a number? (not just the character '1' but the whole number '19' and '20').

char s[80];
strcpy(s,"a 19 b c d 20");

int i=0;
int num=0;
int digit=0;
for (i =0;i<strlen(s);i++){
    if ((s[i] <= '9') && (s[i] >= '0')){    //how do i test for the whole integer value not just a digit

        //if number then convert to integer
        digit = s[i]-48;
        num = num*10+digit;
    }

    if (s[i] == ' '){
        break; //is this correct here? do nothing
    }
    if (s[i] == 'a'){
       //copy into a temp char
    }
}

Answer 1

You can use atoi().

Answer 2

after your if you need to shift to while to collect subqsequent digits until you hit a non-digit.

BUT, more inportantly, have you clearly defined your requirements? Will you allow whitespace between the digits? What if there are two numbers, like abc123def456gh?

Answer 3

There's no way to test for a whole number. Writing a lexer, as you've done is one way to go. Another would be to try and use the C standard library's strtoul function (or some similar function depending on whether the string has floating point numbers etc).

Your code needs to allow for whitespaces and you can use the C library's isdigit to test if the current character is a digit or not:

vector<int> parse(string const& s) {
     vector<int> vi;
     for (size_t i = 0; i < s.length();) {
         while (::isspace((unsigned char)s[ i ]) i++;
         if (::isdigit((unsigned char)s[ i ])) {
              int num = s[ i ] - '0';
              while (::isdigit((unsigned char)s[ i ])) {
                 num = num * 10 + (s[ i ] - '0');
                 ++i;
              }
              vi.push_back(num);
         }
                ....

Another approach will be to use boost::lexical_cast:

 vector<string> tokenize(string const& input) {
     vector<string> tokens;
     size_t off = 0, start = 0;
     while ((off = input.find(' ', start)) != string::npos) {
          tokens.push_back(input.substr(start, off-start));
          start = off + 1;
     }
     return tokens;
 }

 vector<int> getint(vector<string> tokens) {
      vector<int> vi;
      for (vector<string> b = tokens.begin(), e = tokens.end(); b! = e; ++b) {
         try
         {
           tokens.push_back(lexical_cast<short>(*b));
         }
         catch(bad_lexical_cast &) {}
      }
      return vi;
 }

Answer 4

These are C solutions:

Are you just trying to parse the numbers out of the string? Then you can just walk the string using strtol().

long num = 0;
char *endptr = NULL;
while (*s) {
  num = strtol(s, &endptr, 10);
  if (endptr == s) { // Not a number here, move on.
    s++;
    continue;
  }
  // Found a number and it is in num. Move to next location.
  s = endptr;
  // Do something with num.
}

If you have a specific location and number to check for you can still do something similar.
For example: Is '19' at position 10?

int pos = 10;
int value = 19;
if (pos >= strlen(s))
  return false;
if (value == strtol(s + pos, &endptr, 10) && endptr != s + pos)
  return true;
return false;

Are you trying to parse out the numbers without using any library routines?

Note: I haven't tested this...

int num=0;
int sign=1;
while (*s) {
  // This could be done with an if, too.
  switch (*s) {
    case '-':
      sign = -1;
    case '+':
      s++;
      if (*s < '0' || *s > '9') {
        sign = 1;
        break;
      }
    case '0':
    case '1':
    case '2':
    case '3':
    case '4':
    case '5':
    case '6':
    case '7':
    case '8':
    case '9':
      // Parse number, start with zero.
      num = 0;
      do {
        num = (num * 10) + (*s - '0');
        s++;
      } while (*s >= '0' && *s <= '9');
      num *= sign;
      // Restore sign, just in case
      sign = 1;
      // Do something with num.
      break;
    default:
      // Not a number
      s++;
  }
}

Answer 5

Its not very clear what you are looking for.. Assuming you want to extract all the digits from a string and then from a whole number from the found digits you can try the following:

    int i;
    unsigned long num=0; // to hold the whole number.
    int digit;
    for (i =0;i<s[i];i++){
            // see if the ith char is a digit..if yes extract consecutive digits
            while(isdigit(s[i])) { 
                    num = num * 10 + (s[i] - '0');
                    i++;
            }
    }

It is assumed that all the digits in your string when concatenated to from the whole number will not overflow the long data type.

Answer 6

It seems like you want to parse the string and extract all the numbers from it; if so, here's a more "C++" way to do it:

string s = "a 19 b c d 20"; // your char array will work fine here too
istringstream buffer(s);
string token;
int num;
while (!buffer.eof())
{
    buffer >> num; // Try to read a number
    if (!buffer.fail()) {    // if it doesn't work, failbit is set
        cout << num << endl; // It's a number, do what you want here
    } else {
        buffer.clear();  // wasn't a number, clear the failbit
        buffer >> token; // pull out the non-numeric token
    }
}

This should print out the following:

19 
20

The stream extraction operator pulls out space-delimited tokens automatically, so you're saved from having to do any messy character-level operations or manual integer conversion. You'll need to #include <sstream> for the stringstream class.