Calculate exit code with $? in shell

Question

Could you guys help to give the algorithm used to generate $? in shell from exit code in program? For example,

  $? is 1 for exit(1);
  $? is 255 for exit(-1);

So I can infer exit code from $?

  $? is 1 => exit code is 1
  $? is 255 => exit code is -1

For something special,

  $? is 0 for exit(256);
  $? is 1 for exit(257);

Could anyone give the algorithm in shell used to generate $? with exit code, so that I can know the exit code only by observing $? after executing a command.

Thanks a lot.

Edit: To answer the question below, I added this example.

----a.c----
1 #include <stdlib.h>
2 int main()
3 {
4   exit(-1);
5 }
ning@m:~/work/02_test/ctest> gcc a.c
ning@m:~/work/02_test/ctest> ./a.out 
ning@m:~/work/02_test/ctest> echo $?
255
ning@m:~/work/02_test/ctest>

Answer 1

With bash, $? is the exit code of the last command. Running a shell script .

#!/bin/bash
exit 113

then echo $? shows 113. See http://tldp.org/LDP/abs/html/exit-status.html for more info

With BASH there are some common exit codes.

The return value is just an 8bit Int. Exit codes 255 means out of range, so -1. The reason you are seeing 256->0 and 257->1 is it's wrapping around. Basically the exit value modulo 256.

All of this is true with Bash or C