It's a string rather than a integer because substr returns a strings, not integers.

You say Perl is typeless when the whole issue you're having is that you think that substr returns a value of a different type than it does.

You seem to claim your approach would work in C, but it wouldn't. 

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

char* substr(const char* src, size_t start, size_t len) {
   while (start && *src) {
      ++start;
      ++src;
   }

   char* dst = malloc(len+1);
   if (!dst)
      return NULL;

   dst[len] = 0;
   return strncpy(dst, src, len);
}

int main() {
   const char* s = "\x90";
   char* first = substr(s, 0, 1);
   printf("%02X\n", first);
   free(first);
   return 0;
}

[download]
Output: 
$ gcc -Wall -Wextra -pedantic -std=c99 a.c -o a && a
a.c: In function ‘main’:
a.c:22:4: warning: format ‘%X’ expects argument of type ‘unsigned int’
+, but argument 2 has type ‘char *’ [-Wformat=]
    printf("%02X\n", first);
    ^
1E25970

[download]

You need to do something to extract the number from the string: *first (C) or ord($first) (Perl).

In reply to Re^5: I've muddled my bit and byte formats. by ikegami
in thread I've muddled my bit and byte formats. by murrayn

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