I want it to ouput "abcd", but it dosen't.

  1 #include<stdio.h>
  2 int main()
  3 {
  4         printf("a\x62cd");
  5 }
  6

printf("a\x62""cd");

That's what I ended up having to do before. That, or seperating it out of the string entirely as in printf("a%ccd", 0x62);

Why don't the complier just translates the fixed few of characters(eg: two char) after \x to an escape char?

Why should it? There are systems where "char" isn't 8 bits... Or were, decades ago, and the C standard people latch onto this technicality like a bear trap and refuse to let go.

It plainly doesn't assume it stops at 8 bits, anyway, so the question's kind of pointless.

Unicode characters aren't 8 bits, and they've been spotted in use somewhat more recently than decades ago.

The only way to produce a deterministic result when the number of output bits can not be constrained is with a maximal-munch parser.

I tried your code with the same result as you. Then I just tried the following code:

  1 #include<stdio.h>
  2 int main()
  3 {
  4         printf("a\x62mn");
  5 }
  6

#rob@fred:~/programs$./unixcom.exee
abmn
rob@fred:~/programs$

achenle - unicode or utf-16 or whatever else - have been around for a long time. They are not considered 'char' in C, they are wchar, wide characters, a different datatype. This datatype affects file orientation, which is the primary way reads & writes occur on tyy/file/device(s). wchar are NOT char, by definition and practice

try man wchar

And there still are embedded systems with 32 bit char.