Counting number of occurrences of a char in a string in C

Question

I have the string str

char *str = "100.10b.100.100"; 

I want to count the occurrences of '.' in str, preferably a one-liner. (If possible no loops)

My approach would be the standard strchr:

  int i = 0;   char *pch=strchr(str,'.');   while (pch!=NULL) {     i++;     pch=strchr(pch+1,'.');   } 

Answer 1

Here's the way I'd do it (minimal number of variables needed):

for (i=0; s[i]; s[i]=='.' ? i++ : *s++); 

Answer 2

OK, a non-loop implementation (and yes, it is meant as a joke).

size_t CountChars(const char *s, char c) {   size_t nCount=0;   if (s[0])   {     nCount += ( s[0]==c);     if (s[1])     {       nCount += ( s[1]==c);       if (s[2])       {         nCount += ( s[2]==c);         if (s[3])         {           nCount += ( s[3]==c);           if (s[4])           {             nCount += ( s[4]==c);             if (s[5])             {               nCount += ( s[5]==c);               if (s[6])               {                 nCount += ( s[6]==c);                 if (s[7])                 {                   nCount += ( s[7]==c);                   if (s[8])                   {                     nCount += ( s[8]==c);                     if (s[9])                     {                       nCount += ( s[9]==c);                       if (s[10])                       {                         /* too long */                         assert(0);                       }                     }                   }                 }               }             }           }         }       }     }   }   return nCount; }