Confused about stdin stdout stderr? [closed] - unix

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So I understand the meaning of the command such as 1>&2 or 2>&1 but
my question is why do we use such commands? Redirecting the stdout to stderr or from stderr to stdout?
Also for $[$1 $sign $2] command (used for simple calculator), is $sign a default command in Unix? What does it mean? What about the $ that is outside of the brackets?

Answer 1

Two questions for the price of one. It would be better asked as two questions since they are almost wholly unrelated.
I/O Redirection
Why would you use 1>&2?
A standard reason is so a shell script reports errors on stderr instead of stdout. For example:
if [ ! -f "$file" ]
then
echo "$0: no such file - $file" >&2
exit 1
fi
Why would you use 2>&1?
You want to capture all of the output from a command. For example, you might be about to run a find command that will take ages, so you might write:
find $HOME -name '*perq*' -print > log.file 2>&1 &
All the data and error messages from find are sent to the log.file which you can study later when the command is complete.
The [ (test) command
You ask about $[$1 $sign $2].
This has so many problems it is difficult to answer coherently. The question about a calculator suggests that maybe $1 is supposed to be a number (let's use 13) and $2 is too (let's use 9), and $sign might be either + or -.
$[ is not normally defined as a variable, so the shell will try to find a command $[13 on the PATH and (normally) will fail.
If you had written [ $1 $sign $2 ], then we'd be better off. There is a command [ (also known as test). It is usually a shell built-in, but originally was a separate executable (and you usually find that there is a /usr/bin/test (or /bin/test) and /usr/bin/[ (or /bin/[) executable even on modern systems. Note that like all commands, the name ([) is separated from its arguments by spaces. The [ command requires the last argument to be ]. Now the expression might be:
[ 13 + 9 ]
[ 13 - 9 ]
However, the test command would object; it doesn't support arithmetic. It does support string comparisons = and !=; it supports numeric comparisons with -eq, -ne, -lt, -le, -gt or -ge.
If you want to do arithmetic, you either use the expr command, or one of the built-in alternatives:
x=$(($1 $sign $2))
This will assign 22 or 4 (depending on the value of $sign) to the variable x.

Answer 2

answer for redirection :
In unix the data/text file's are internally associated with something known as file handlers (integer's value) which help the OS distinguish and identify a file during processing. everything in unix is considered to be a file , even the input and output devices.
standard input is the keyboard , the standard output is stdout ie monitor which also functions as the stderr file stream , all this is by default
hence these are associated as file's hence have numbers assigned to them ....0 ,1 and 2.
so when you use something like 2>&1 you are basically telling the OS to put the stderr(2) data into the stdout(1) data stream
answer for $ sign :
the dollar sign here is a way to access the value of the variables.
example when you have a variable a like as below ::
Nitin#Kaizen > a=5
Nitin#Kaizen > echo a --> will just print a
Nitin#Kaizen > echo $a --> will print 5
note : $a is same as ${a} or $[a] , they access the value of a variable .... kind of de-referencing
$[$1 $sign $2] :: you issue statement
Nitin#Kaizen > a=5
Nitin#Kaizen > b=3
Nitin#Kaizen > sign=+
Nitin#Kaizen > echo `expr $[ $a $sign $b ]`
8 --- output is same as 5 + 3 , the value of a + b
answer for why you are confused :
you need make a slight effort to read a book completely again. these are very basics and hence become clear with a bit of programing and study.
Any good unix basic book would do that covers the chapters on File System (UFS) and a bit on scripting will do !!
hope this helps.