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I understand that if I execute a command in the shell with the ampersand operators e.g. command1&&command2 and should the first command execute successfully it does so with a zero exit status.

Is this similar or different to that of boolean logic whereby 0 is false and 1 is true or have I understood it incorrectly?

Does the type of shell you use make a difference? I am currently running the Bourne-Again Shell.

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2 Answers 2

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Shell mostly is 0 to indicate zero errors and success, so yes in this sense it is true logically, unlike in many programming languages where 0 usually is false. Although modern languages tend to have distinct logical values of true and false these days.

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All the shells I know. I said mostly because I do not know all the shells. –  johnshen64 Apr 30 '12 at 2:20
    
Thanks. How does this differ to a non-zero exit status for example when using control operators such as || in command such as command1||command2? –  PeanutsMonkey Apr 30 '12 at 2:21
    
It means if command1 fails then execute command2, that is command1 exit status is nonzero. –  johnshen64 Apr 30 '12 at 2:22
    
Wonder why they chose the opposite representation of boolean values. –  PeanutsMonkey Apr 30 '12 at 2:55
    
Here is a good explanation, ry.ca/blog/category/tech/unix but as to why only mr. Bourne who developed the original grandfather of all shells would know. I guess it just means 0 or no error, thus success, thus true. –  johnshen64 Apr 30 '12 at 3:41

The better rationale is that there is really only one 'success', but there are multiple reasons for a 'failure': file not found, argument error, computational error, etc. This was a convention of the OS, not just of the shell, in the interprocess communications - the return code sent by the exit(2) system call and received by the wait(2) call.

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