read num1
2
read num2
5
echo "$((num1+num2))"
7
echo "$(($num1+$num2))"
7
I referred this.
I don't understand how the above both expressions yields same result? Please explain.
echo "$[num1+num2]"
It doesn't work as suggested in that question. Am i doing anything wrong?
echo 'expr "$num1" + "$num2"`
also not working. Please provide some suggestions.
To evaluate an arithmetic expression, the shell first expands variable and command substitutions inside it. For example, in echo "$(($num1+$num2))", the first thing that happens is that $num1 and $num2 are replaced by the variables' values. The expression becomes 2+5. This is parsed as an arithmetic expression and evaluated to the number 7. The result of parsing and expanding "$(($num1+$num2))" is thus 7.
Arithmetic expressions can contain variable names. These are evaluated to the value of the variable. Thus $((num1+num2)) is the arithmetic expression num1+num2 (whereas $(($num1+$num2)) is the arithmetic expression 2+5). In this case the result is the same.
The two expressions are different when the value of num1 or num2 isn't just a sequence of digits. For example, consider the following snippet:
a="1+2"
echo "$(($a * 4))"
echo "$((a * 4))"
In the second line, the arithmetic expression to be evaluated is 1+2 * 4. This is parsed as 1 + (2 × 4) since multiplication has a higher precedence than addition. The fact that the plus sign came from the expansion of a variable is forgotten by the time the arithmetic evaluation takes place. The result is the number 9, so the first call to echo prints 9.
The behavior of the third line depends on the shell. Some shells (e.g. dash) complain that a does not contain a valid numeric value. Other shells (e.g. ksh, bash, zsh) treat the value of a as a subexpression which is evaluated to get the value to use in the main expression. That is, in these other shells, the value of a is evaluated to 3, and the arithmetic expression calculates 3 × 4 so the result is 12.
Finally, $[…] is a deprecated, non-standard variant of $((…)). The expr utility is an older way of making arithmetic calculations in shell scripts, dating back from the time when shells didn't have a built-in arithmetic syntax. expr "$num1" + "$num2" prints 7, as does the redundant echo `expr "$num1" + "$num2"` (you mistyped one of the quotes).
In POSIX shell, Arithmetic Expansion has form:
$((expression))
If expression contains variables, and those variables contain valid integer value - leading plus or minus is fine - then "$((var))" and "$(($var))" will return the same result (Note: using unsanitized data in Shell Arithmetic evaluation leads to security implication).
The form $[expression] is not defined by POSIX, is deprecated and will be remove in next version of bash.
The expr utility can be used to do arithmetic. In your case, you used wrong syntax for command substitution, it's a backstick, not a single quote:
echo `expr "$num1" + "$num2"`
But you should use "$()" form instead.
