Odtnhj

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 != ’’) 
 if ( *p < ’0’ 
 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; 

// 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 != ’’) *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.