Is there a way to count tokens in C?

Question

I\'m using strtok to split a string into tokens. Does anyone know any function which actually counts the number of tokens?

I have a command string and I nee

Answer 1

One approach would be to simply use strtok with a counter. However, that will modify the original string.

Another approach is to use strchr in a loop, like so:

int count = 0;
char *ptr = s;
while((ptr = strchr(ptr, ' ')) != NULL) {
    count++;
    ptr++;
}

If you have multiple delimiters, use strpbrk:

while((ptr = strpbrk(ptr, " \t")) != NULL) ...

Answer 2

Here is a version based on strtok which does not modify the original string, but a temporal copy of it. This version works for any combination of tabs and space characters used as tokens separators. The function is

unsigned long int getNofTokens(const char* string){
  char* stringCopy;
  unsigned long int stringLength;
  unsigned long int count = 0;

  stringLength = (unsigned)strlen(string);
  stringCopy = malloc((stringLength+1)*sizeof(char));
  strcpy(stringCopy,string);

  if( strtok(stringCopy, " \t") != NULL){
    count++;
    while( strtok(NULL," \t") != NULL )
        count++;
  }

  free(stringCopy);
  return count;
}

A function call could be

char stringExample[]=" wordA 25.4 \t 5.6e-3\t\twordB 4.5e005\t ";
printf("number of elements in stringExample is %lu",getNofTokens(stringExample));

Output is

number of elements in stringExample is 5

Answer 3

As number of tokens is nothing but one more than the frequency of occurrence of the delimiter used. So your question boils down to find no. of times of occurrence of a character in a string

say the delimiter used in strtok function in c is ' '

int count =0,i;
char str[20] = "some string here";

for(i=0;i<strlen(str);i++){
    if(str[i] == ' ')
        count++;
}

No. of tokens would be same as count+1