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I have an associative operation >>. The problem is that its cost linearly depends on the size of its left operand. So an expression formed by a sequence of n applications of >> like
a >> a >> a >> a >> a >> ... >> a it has quadratic cost in terms of n, because by default infix operators are left-associative. How to make it right-associative so that the cost of such an expression is kept linear in terms of n?
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The right-associativity of the = operator allows expressions such as a = b = c to be interpreted as a = (b = c) . In C++, the assignment a = b is an expression that evaluates to the same value as the expression a , with the side effect of storing the R-value of b into the L-value of a .
Operators Associativity is used when two operators of same precedence appear in an expression. Associativity can be either Left to Right or Right to Left. For example: '*' and '/' have same precedence and their associativity is Left to Right, so the expression “100 / 10 * 10” is treated as “(100 / 10) * 10”.
The grouping of operands can be forced by using parentheses. For example, in the following statements, the value of 5 is assigned to both a and b because of the right-to-left associativity of the = operator.
I found a solution. Scala reference says in section 6.12.3 Infix Operations:
The associativity of an operator is determined by the operator’s last character. Operators ending in a colon ‘:’ are right-associative. All other operators are left-associative.
Therefore it was enough to rename >> to >>:.
It took me some time to realize that while a >> b is desugared into a.>>(b), a >>: b is desugared into b.>>:(a). So I had to define >>: as
def >>:(x: T): T = x >> this If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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