c - Pointer De-referencing -

#include<stdlib.h> #include<stdio.h> #define no_of_chars 256  /* returns array of size 256 containg count    of characters in passed char array */ int *getcharcountarray(char *str) {    int *count = (int *)calloc(sizeof(int), no_of_chars);    int i;    (i = 0; *(str+i);  i++)       count[*(str+i)]++;    return count; }  /* function returns index of first non-repeating    character in string. if characters repeating     returns -1 */ int firstnonrepeating(char *str) {   int *count = getcharcountarray(str);   int index = -1, i;    (i = 0; *(str+i);  i++)   {     if (count[*(str+i)] == 1)     {       index = i;       break;     }      }      free(count); // avoid memory leak   return index; }  /* driver program test above function */ int main() {   char str[] = "geeksforgeeks";   int index =  firstnonrepeating(str);   if (index == -1)       printf("either characters repeating or string empty");   else    printf("first non-repeating character %c", str[index]);   getchar();   return 0; } 

i can't grasp following lines:

count[*(str+i)]++; 

amd

  int *getcharcountarray(char *str)     {        int *count = (int *)calloc(sizeof(int), no_of_chars);        int i;        (i = 0; *(str+i);  i++)           count[*(str+i)]++;        return count;     } 

the program used find first non-repeating character in string.

i can't grasp following lines:

count[*(str+i)]++; 

work outside in:

• since str pointer char , i int, str + i pointer char i chars after 1 str points to
• *(str+i) dereferences pointer str+i, meaning evaluates char pointer points to. equivalent str[i].
• count[*(str+i)] uses char @ index i in string str index dynamic array count. expression designates int @ index (since count points array of ints). see below.
• count[*(str+i)]++ evaluates int @ index *(str+i) in array count points to. side effect, increments array element 1 after value of determined expression. overall expression present in code exclusively side effect.

it important note although space reserved in array count counting appearances of 256 distinct char values, expression asked is not safe way count of them. that's because type char can implemented signed type (at c implementer's discretion), , common implemented way. in case, non-negative char values correspond array elements, , undefined behavior result if input string contains others. safer be:

#include <stdint.h>  # ...  count[(uint8_t) *(str+i)]++; 

i.e. same original, except explicitly casting each character of input string unsigned 8-bit value.

overall, function creates array of 256 ints, 1 each possible char value, , scans string count number of occurrences of each char value appears in it. returns array of occurrence counts.