- bash - What are the special dollar sign shell variables . . . - Stack . . .
In Bash, there appear to be several variables which hold special, consistently-meaning values For instance, myprogram amp;; echo $! will return the PID of the process which backgrounded myprog
- Whats the difference between lt; lt;, lt; lt; lt; and lt; lt; in bash?
Here-strings in bash are implemented via temporary files, usually in the format tmp sh-thd <random string>, which are later unlinked, thus making them occupy some memory space temporarily but not show up in the list of tmp directory entries, and effectively exist as anonymous files, which may still be referenced via file descriptor by the
- bash - What is the purpose of in a shell command? - Stack Overflow
Furthermore, you also have which is the logical or, and also which is just a separator which doesn't care what happend to the command before
- shell - Bash regex =~ operator - Stack Overflow
What is the operator =~ called? I'm not sure it has a name The bash documentation just calls it the =~ operator Is it only used to compare the right side against the left side? The right side is considered an extended regular expression If the left side matches, the operator returns 0, and 1 otherwise Why are double square brackets required when running a test? Because =~ is an operator of
- bash - Shell equality operators (=, ==, -eq) - Stack Overflow
The reason is that Bash is untyped The -eq causes the strings to be interpreted as integers if possible including base conversion: And 0 if Bash thinks it is just a string: So [[ "yes" -eq "no" ]] is equivalent to [[ 0 -eq 0 ]] Last note: Many of the Bash specific extensions to the Test Constructs are not POSIX and therefore may fail
- What does $# mean in bash? - Ask Ubuntu
Furthermore, when you use bash -c, behavior is different than if you run an executable shell script, because in the latter case the argument with index 0 is the shell command used to invoke it
- bash - Precedence of the shell logical operators , || - Unix Linux . . .
I am trying to understand how the logical operator precedence works in bash For example, I would have expected, that the following command does not echo anything true || echo aaa amp; amp; echo
- bash - Difference between gt; gt; and - Unix Linux Stack Exchange
In general, in bash and other shells, you escape special characters using \ So, when you use echo foo >\> what you are saying is "redirect to a file called > ", but that is because you are escaping the second > It is equivalent to using echo foo > \> which is the same as echo foo > '>' So, yes, as Sirex said, that is likely a typo in your book
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