Re^2: why divide 256 for the exit code

thanks monk, for example, if i write a C program as below: main() { exit(100); } then, i invoke it from perl, and print the exit code, it's 25600. it convert to binary is: 1100100 00000000. so the high 8 bits is: 1100100(100), and the low 8 bits is: 00000000. if this C programm ran abnormally, the low 8 bits may contain info of interest, so use $? & 127 OR $? & 128. am'i right?

Comment on Re^2: why divide 256 for the exit code

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Re^3: why divide 256 for the exit code by Illuminatus (Curate) on Sep 07, 2009 at 16:12 UTC
You want 8 bits, not 7. IMHO, if you are doing bit operations, you should use hex (or octal if you are old-school) representations, so that it is clear. In this case, you should & with 0xff	[reply]
Re^4: why divide 256 for the exit code by almut (Canon) on Sep 07, 2009 at 16:32 UTC
In this case, you should & with 0xff Actually, I think it's sufficient to use `$? & 0x7f` (or `$? & 127` in decimal) to test for abnormal termination. That's what the code snippet in system suggests. Also, the system header files responsible for the exit status handling (e.g. `/usr/include/bits/waitstatus.h` in my case) usually define those macros (among others): `#define __WTERMSIG(status) ((status) & 0x7f) /* Nonzero if STATUS indicates normal termination. */ #define __WIFEXITED(status) (__WTERMSIG(status) == 0)` [download] (from that I would conclude that the core-dumped bit (`$? & 128`) doesn't occur in isolation (i.e. without WTERMSIG being non-zero) )	[reply] [d/l] [select]