C 将 char* 字符串转换为 int 数组

Question

I have a char* string of only exactly 5 digits. I want to convert this string to a integer array.

I've tried this:

#include <stdio.h>
#include <string.h>

int main()
{
     int numbers[5];
     const char* titleid = "TEST00411";
     const char* digits = titleid + 4;
     
     for (int i = 0; i < 5; ++i) {
          numbers[i] = digits[i];
          printf("LOOP: %d\n", digits[i]);
     }
     
     printf("%d\n", numbers[0]);
     printf("%d\n", numbers[1]);
     printf("%d\n", numbers[2]);
     printf("%d\n", numbers[3]);
     printf("%d\n", numbers[4]);
     
     return 0;
}

Output:

LOOP: 48
LOOP: 48
LOOP: 52
LOOP: 49
LOOP: 49
48
48
52
49
49

Why aren't the numbers displayed correctly (0, 0, 4, 1, 1)?

Answer 1

What you are getting is the ASCII equivalent characters. Subtract '0' to get the original number from ASCII characters.
Here is the code to just do it.

for (int i = 0; i < 5; ++i) {
          numbers[i] = digits[i]-'0';
          printf("LOOP: %d\n", digits[i]);
}