How to understand if condition in bash? - bash

Question

I am a little confused about Bash test expression and if test commands:
(1)
#!/bin/bash
count=90
if [ $((count)) ]; then
echo "True "
else
echo "False"
fi
Executing it:
# ./test1.sh
True
(2)
#!/bin/bash
count=90
if $((count)); then
echo "True "
else
echo "False"
fi
Executing it:
# ./test2.sh
./test2.sh: line 5: 90: command not found
False
(3)
#!/bin/bash
count=90
if ((count)); then
echo "True "
else
echo "False"
fi
Executing it:
# ./test3.sh
True
My questions are as follows:
a) For test2.sh, why does it complain "command not found"?
b) Although test1.sh and test3.sh outputs the same result, do they have same meanings in if condition?

Answer 1

Anything that does not produce an error is valid.
That said, there are conventions that people follow when writing shell scripts, and they exist mostly because they make sense.
The if command is a construct that:
runs a command, then on the basis of the exit code of that command,
runs some other command.
For example, your system has programs named true and false on it. So you can make an if construct like:
if true; then
echo TRUE
else
echo FALSE
fi
When you "wrap a condition in square brackets" in shell, what you're really doing is running a command named [. It may be a built-in in your shell, or it may be located at /bin/[, but it's a command either way. Its options appear like conditions, and its purpose is to produce an exit value that will be consumed by the if command.
Now ... when you do arithmetic in bash, you can use constructs like $((...)) which is called "Arithmetic Expansion" because the result of the arithmetic is expanded to replace the expression, as if it were a variable. When you use ((...)), without the preceding dollar sign, then the expression is simply evaluated, rather than printed.
So .. Your first if command tests to see that the arithmetic expansion evaluates to true. Most valid arithmetic should do this. Your second command executes the expansion as if it were a command, which is not. And you get the error telling you that 90 can't be run as a command. And your third command executes the arithmetic, but without expanding it. As with the first option, as long as the arithmetic is valid, the test returns true.
The difference between the first and the third variants is that in the first case, the test command (a.k.a. /bin/[ or your shell's built-in equivalent) is evaluating the results of your arithmetic, the result of which is always true unless you do something silly like try to use decimal numbers, whereas in the third case, a valid arithmetic expression that results in a 0 (zero) will appear to be "false".
To test this difference in your shell, try the following:
$ if [ $(( 2.5 + 2 )) ]; then echo yes; else echo no; fi # ERROR
$ if (( 2.5 + 2 )); then echo yes; else echo no; fi # ERROR, no
$ if [ $(( 2 + 2 )) ]; then echo yes; else echo no; fi # yes
$ if (( 2 + 2 )); then echo yes; else echo no; fi # yes
$ if [ $(( 2 - 2 )) ]; then echo yes; else echo no; fi # yes
$ if (( 2 - 2 )); then echo yes; else echo no; fi # no
Either behaviour may be what you're looking for, but you haven't indicated what problem you're trying to solve, so I can't recommend one over the other.