Why do I prefer the format specifier %d over %i in printf?

You can also use %i for printing integer numbers, which may sound more intuitive, because it begins with i. Most C code uses %d however, because in scanf, %d and %i have a subtle difference. %d only reads decimal numbers, but %i can read other bases like octal and hexadecimal as well. So even %i and %d have the same meaning in printf, I would use %d for consistency between scanf and printf format strings.

#include <stdio.h>
#define INPUT_STRING_SIZE 10

void parse(char parsed_string[], char format_specifier[]) {
  int number;
  int parsed_argument_count = sscanf(parsed_string, format_specifier, &number);
  // scanf reads from the keyboard, but sscanf reads from a string.
  // We use sscanf to avoid keyboard entry
  printf("Parsing %s using %s\n", parsed_string, format_specifier);
  printf("Number of successfully parsed arguments: %d\n",
         parsed_argument_count);
  printf("Read data as decimal integer: %d\n", number);
  puts("");
}

int main() {
  parse("-0xf", "%d");
  parse("-0xf", "%i");
  parse("-0b1111", "%d");
  parse("-0b1111", "%i");
}

Parsing -0xf using %d
Number of successfully parsed arguments: 1
Read data as decimal integer: 0

Parsing -0xf using %i
Number of successfully parsed arguments: 1
Read data as decimal integer: -15

Parsing -0b1111 using %d
Number of successfully parsed arguments: 1
Read data as decimal integer: 0

Parsing -0b1111 using %i
Number of successfully parsed arguments: 1
Read data as decimal integer: -15