() / () (block)

() / ()
() / ()
Category	Operators
Type	Reporter
Introduced in	11Oct03

The () / () block is an Operators Block and a Reporter Block. The block divides the second value from the first and returns the result.

If the first value is not evenly divisible by the second, the reported value will have decimals. To find the remainder instead, use the () Mod () block.

The numbers can be typed directly into the block, or Reporter blocks can be used instead.

This block can be stacked inside itself — this can be used to fit more numbers in.

Note:

Dividing by 0 in the online editor will return infinity, NaN, or negative infinity. It depends on if the numerator is positive, 0, or negative; respectively. This is sometimes used with other blocks to create new possible changes to a project. If attempted on the 1.4 offline editor, it will give a Script Error and stops the script.

Example Uses

In many projects, numbers must be divided — this block will do the job.

Some common uses for the () / () block are as follows:

• Calculator scripts

if <(operation) = [division]> then
set [answer v] to ((input1) / (input2))
end

• Dividing lists of numbers

set [i v] to (1)
repeat (length of [list v])
replace item (i) of [list v] with (round ((item (i) of [list v]) / (2)))

• Mathematical formulas

set [area v] to (((base) * (height)) / (2)) //area of a triangle

Workaround

Main article: List of Block Workarounds

The block can be replicated with the following code:

set [counter v] to [0]
set [dividend v] to []
set [decimal point v] to [0]
repeat (length of (a))
    change [counter v] by (1)
    if <(letter (counter) of (a)) = [.]> then
        set [decimal point 1 v] to ((length of (a)) - (counter))
    else
        if <not <(letter (counter) of (a)) = [-]>> then
            set [dividend v] to (join(dividend)(letter (counter) of (a)))
        end
    end
end
set [counter v] to [0]
set [divisor v] to [0]
set [decimal point 2 v] to [0]
repeat (length of (b))
    change [counter v] by (1)
    if <(letter (counter) of (b)) = [.]> then
        set [decimal point 2 v] to ((length of (b)) - (counter))
    else
        if <not <(letter (counter) of (b)) = [-]>> then
            set [divisor v] to (join(disivor)(letter (counter) of (b)))
        end
    end
end
set [remainder v] to []
set [calculating v] to []
set [conuter v] to [0]
repeat (. . .:: grey) // set precision
    change [counter v] by (1)
    if <(length of (dividend)) < (counter)> then
        set [remainder v] to (join(remainder)(0))
    else
        set [remainder v] to (join(remainder)(letter (counter) of (dividend)))
    end
    set [greatest factor v] to [0]
    repeat until <(remainder) < ((divisor) * (greatest factor))>
        change [greatest factor v] by (1)
    end
    if <(remainder) < ((divisor) * (greatest factor))> then
        change [greatest factor v] by (-1)
    end
    set [calculating v] to (join(calculating)(greatest factor))
    set [remainder v] to ((remainder) - ((greatest factor) * (divisor)))
end
set [counter v] to [0]
set [quotient v] to []
repeat (length of (calculating))
    change [counter v] by (1)
    set [quotient v] to (join(quotient)(letter(counter) of (calculating)))
    if <(((length of (dividend)) - (decimal point 1)) + (decimal point 2)) = (length of (quotient))> then
        set [quotient v] to (join(quotient)[.])
    end
end
if <not<<(a) < [0]> = <(b) < [0]>>> then
    set [quotient v] to (join[-](quotient))
end

The (quotient) variable will contain the quotient.

A more concise workaround is this:

if <(b) < (0)> then
set [result v] to (((a) * ([e^ v] of ((-1) * ([ln v] of ((-1) * (b)))):: operators)) * (-1))
else
set [result v] to ((a) * ([e^ v] of ((-1) * ([ln v] of (b))):: operators))
end

How it Works

Let's look at the expression in the "else" part of the statement. This is the expression:

$a\times {e^{-1\times \ln {b}}}={\frac {a}{b}}$

divided both sides by "a" and take the natural log of both sides to get:

$-1\times \ln {b}=\ln(b^{-1})$

The negative before a log can be rewritten as an exponent in the log, like this:

$a\times \log {b}=\log {b^{a}}$

So: $-1\times \ln {b}=\ln(b^{-1})$

Because the log of a negative number does not exist, the formula in the "if" part of the statement pretends like "b" is positive and then just flips the answer.

Because Scratch reports "ln 0" as -Infinity, you can divide by zero using this expression and not get an error.

See Also

• () + ()
• () - ()
• () * ()