I have been trying to understand this code of converting integers from to words from a well known website but I am not getting the pointer arithmetic with respect to the minus sign single_digits[*num - '0'] in it so can anyone please make me understand it? 

/* C program to print a given number in words. The program handles numbers from 0 to 9999 */ #include <stdio.h> #include <string.h> #include <stdlib.h> /* A function that prints given number in words */ void convert_to_words(char *num) { int len = strlen(num); // Get number of digits in given number /* Base cases */ if (len == 0) { fprintf(stderr, "empty string\n"); return; } if (len > 4) { fprintf(stderr, "Length more than 4 is not supported\n"); return; } /* The first string is not used, it is to make array indexing simple */ char *single_digits[] = { "zero", "one", "two", "three", "four","five", "six", "seven", "eight", "nine"}; /* The first string is not used, it is to make array indexing simple */ char *two_digits[] = {"", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"}; /* The first two string are not used, they are to make array indexing simple*/ char *tens_multiple[] = {"", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"}; char *tens_power[] = {"hundred", "thousand"}; /* Used for debugging purpose only */ printf("\n%s: ", num); /* For single digit number */ if (len == 1) { printf("%s\n", single_digits[*num - '0']); return; } /* Iterate while num is not '\0' */ while (*num != '\0') { /* Code path for first 2 digits */ if (len >= 3) { if (*num -'0' != 0) { printf("%s ", single_digits[*num - '0']); printf("%s ", tens_power[len-3]); // here len can be 3 or 4 } --len; } /* Code path for last 2 digits */ else { /* Need to explicitly handle 10-19. Sum of the two digits is used as index of "two_digits" array of strings */ if (*num == '1') { int sum = *num - '0' + *(num + 1)- '0'; printf("%s\n", two_digits[sum]); return; } /* Need to explicitely handle 20 */ else if (*num == '2' && *(num + 1) == '0') { printf("twenty\n"); return; } /* Rest of the two digit numbers i.e., 21 to 99 */ else { int i = *num - '0'; printf("%s ", i? tens_multiple[i]: ""); ++num; if (*num != '0') printf("%s ", single_digits[*num - '0']); } } ++num; } } /* Driver program to test above function */ int main(void) { convert_to_words("9923"); convert_to_words("523"); convert_to_words("89"); convert_to_words("8989"); return 0; }
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1 Answer

single_digits[*num - '0']

*num - you get the character referenced by this pointer.

*num - '0' you get the number equal the decimal value of the character (assuming the char is the digit)

single_digits[*num - '0'] you get the array element with index of the decimal value of the character.

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