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Fixed code example which utilizes non-existing function 'abs' - the value is stated to be negative so abs(a) is the same as -a

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Lauri Peltonen 2019-09-13 21:36:43 +03:00
parent 2d601a4f23
commit b7e6dc3d05
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docs/types/value-types.rst
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@ -110,7 +110,7 @@ Modulo
The modulo operation ``a % n`` yields the remainder ``r`` after the division of the operand ``a`` The modulo operation ``a % n`` yields the remainder ``r`` after the division of the operand ``a``
by the operand ``n``, where ``q = int(a / n)`` and ``r = a - (n * q)``. This means that modulo by the operand ``n``, where ``q = int(a / n)`` and ``r = a - (n * q)``. This means that modulo
results in the same sign as its left operand (or zero) and ``a % n == -(abs(a) % n)`` holds for negative ``a``: results in the same sign as its left operand (or zero) and ``a % n == -(-a % n)`` holds for negative ``a``:
* ``int256(5) % int256(2) == int256(1)`` * ``int256(5) % int256(2) == int256(1)``
* ``int256(5) % int256(-2) == int256(1)`` * ``int256(5) % int256(-2) == int256(1)``