20-CS-4003-001 |
Organization of Programming Languages |
Fall 2018 |
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Expression Evaluation |

**Operators**

Scheme: (+ 1 2 3 4) Haskell: (+) (+ 1) 1+2+3+4+5 Java: 1+2+3+4+5 numb.add(new BigInteger("10")) |
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In Scheme all expressions have the form
(op arg arg ...)where op is an operator (also called a function or procedure)
of some sort. This is known as prefix notation. In the
example to the left + is the operator. The return value in
this case is a number.
In Haskell things are different. The first expression on the left returns a procedure which takes two arguments and outputs the sum of those two numbers. It may be invoked like this: ((+) 2 3)which returns 5. The second expression returns a procedure which takes one argument and outputs the sum of the argument plus 1. It may be used like this: ((+ 1) 2)which returns 3. The third expression uses the + operator in
infix form and always returns a number.
In Java, |
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Scheme: (/ 3 (floor (/ 3 4))) Haskell: 2*3+4-2*3 (2*3)+4+(-2*3) tail head [[10,20,30]] tail (head [[10,20,30]]) tail $ head [[10,20,30]] Java: 2*3+4-2*3 import java.util.*; public class a { public static void main (String args[]) { Vector l = new Vector(); Vector x = new Vector(); x.add(10); x.add(20); x.add(30); l.add(x); Vector v = (Vector)l.get(0); v.removeElementAt(0); System.out.println(v); } } |
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In Scheme precedence and associativity is determined by parentheses.
One cannot have a procedure name surrounded by white space unless it
is an argument to a procedure. A nested expression such as the one to
the left is evaluated from innermost parentheses out. Thus, (/ 3
4) is evaluated to 3/4 then (floor 3/4) is
evaluated to 0 and (/ 3 0) raises an exception.
Operator precedence and associativity in Haskell is given by the following (partial) table.
tail is
being applied to two arguments. The fourth expression evaluates to
[20,30]. The $ in the fifth expression changes the
order, evaluating on the right first, and evaluates to [20,30].
In Java, operator precedence is defined in a table similar to that above for
Haskell but with a few changes in order. For non-primitive data types
The order of evaluation is determined by method calls. The example to the
left gives the same result as |
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