Can someone help explain what this C algorithm is doing?

Question

I have been looking at this C code but am not sure what exactly it is doing. I don't understand the use of multiple if statements of finding statements.

int f(char *s) {
  char *p = s;
  int c = 1;
  while (*p == ’ ’)
    ++p;
  while (*p != ’\0’) {
    if ( *p < ’0’ || *p > ’9’ ) {
      printf("Error!\n"); return 0;
    }
  ++p; }
  for (--p; p >= s; --p) {
    if (*p == ’ ’) *p = ’0’;
    *p += c;
    if (*p > ’9’) {
      *p = ’0’; c = 1;
    } else
      c = 0;
    if (c == 0) break;
  }
  if (c != 0) {
    printf("Error!\n");
    return 0;
}
return 1; }

Answer 1

// return an integer given a character pointer, a string.
int f(char *s) {
  // Set current position to start of string
  char *p = s;
  // Initialise carry flag to '1'.
  int c = 1;
  // Move position past leading spaces
  while (*p == ’ ’)
    ++p;
  // Check remaining characters are in the set {'0','1',..,'9'}
  while (*p != ’\0’) {
    // If they are not, return with an error
    if ( *p < ’0’ || *p > ’9’ ) {
      printf("Error!\n"); return 0;
    }
++p; }
  // Now counting back from the end of the string
  for (--p; p >= s; --p) {
    // Turn a space into a '0'; 
    if (*p == ’ ’) *p = ’0’;
    // Increment the digit by the value of the carry; one or zero
    *p += c;
    // This might cause a further carry, capture that
    if (*p > ’9’) {
      *p = ’0’; c = 1;
    } else
      c = 0;
// if no carry, break, else keep on with the carry
    if (c == 0) break;
}
  // If still carrying passed the end of the space, call an error.
  if (c != 0) {
    printf("Error!\n");
    return 0;
}
return 1; }

Essentially: If the input is a digit string, add one; might need a leading space and will use it if the input is all '9's.