At a fundamental level, programming can be viewed as nothing more than applying operators to operands. There is quite a lot of syntactic variation on how this is done:
F -- no parens for zero-operand operators F(X, Y, Z) -- three operands for operator F "+"(X, Y) -- abbreviated X + Y ">"(X, Y) -- abbreviated X > Y
f() // parens required for zero-operand calls f(x, y, z) operator+(x, y) // abbreviated x + y operator>(x, y) // abbreviated x > y
(f) (f x y z) (+ x y) (+ a b c d e) /* Not just binary! */ (+ x (* y z))
&f f x, y, z # Hmm, what does this mean? f(x), y, z # List of three things f(x, y, z) # Three-operand invocation print 1, 3, sort 4, 2 # Gaaack
f () (* ALL ops have 1 operand *) f x y z (* curried *) f(x, y, z) (* uncurried *) op+(x, y) (* same as x + y *)
f // No parens is okay f(x, y, z) // Typical x.+(y) // abbreviated x + y df.format(date) // abbreviated df format date
Operators can be built-in simple things like “+
” or “<
”, or can be built-in or user-defined functions, or can even introduce declarations, modify control flow, and cause side-effects.
Operators have precedence, associativity, arity, and fixity.
Motivation:
Does a • b ¶ c mean ((a • b) ¶ c) or does it mean (a • (b ¶ c)) ?
Higher precedence operators are applied first.
Issues:
Motivation:
Does a • b • c mean ((a • b) • c) or does it mean (a • (b • c)) or are we not even allowed to write such a thing?
We speak of left-associativity, right-associativity, and non-associativity. It is possible, but probably undesirable, for associativity to be unspecified in a language definition 😬.
Issues:
1-5-10
in Python, Smalltalk, and APL.
1-5*10
in Python, Smalltalk, and APL.
1*5-10
in Python, Smalltalk, and APL.
2**3**2
in Python.
-10<-5<-1
in JavaScript, Ruby, Ada, and Python, and explain in detail each of the four completely different behaviors!
The arity of an operator is the allowed number of operands. It can be fixed or variable. A variable arity operator is said to be variadic.
There are a bunch of these: prefix, infix, postfix, overfix, underfix, outfix, and more. Examples in class.
Many language definitions will feature operator tables such as the following:
Operator | Precedence | Associativity | Arity | Fixity |
---|---|---|---|---|
Unary - Unary + | Highest | R | 1 | Prefix |
** | R | 2 | Infix | |
* / % | L | 2 | Infix | |
+ - | L | 2 | Infix | |
< <= == != >= > | NO! | 2 | Infix | |
and or | L | 2 | Infix | |
:= | Lowest | R | 2 | Infix |
The content of these tables vary surprisingly among different languages. It’s nice to have such tables so you can see at a glance how the language arranges its operators.
Oh, and there is always this old bit of advice: When it doubt, just use lots of parentheses.
When there are subexpressions in a complex expression, must certain subexpressions be evaluated before others? Or can they be evaluated in an arbitrary order? Or even in parallel? Would it even matter?
Java forces a left-to-right ordering: a-f(b)-c*d
means do the following, one after another:
time ① | t0 ← f(b)
|
time ② | t1 ← a - t0
|
time ③ | t2 ← c * d
|
time ④ | t3 ← t1 - t2
|
Most languages allow the evaluation order to be undefined so that the compiler can choose the best order it can. This is especially important for parallel architectures (multiprocessor or multicore).
For example, a–f(b)–c*d
can be parallelized as
time ① | t0 ← f(b) |
Also, a:=b[i]; c:=a*2+d*3
can be done like this:
time ① | t0 ← b + i | t1 ← d * 3
|
time ② | a ← *t0 | |
time ③ | t2 ← a * 2 | |
time ④ | c ← t1 + t2 |
Undefined ordering can lead to ambiguities or errors:
f(a,g(b))
could yield different results depending on which order the arguments were evaluated.
a:=b; c:=d
in parallel, but if a and d are aliases of each other we can't.
(a+b)-c
might behave very differently from (a-c)+b
, so a compiler shouldn’t rearrange operands unless it understands these semantics.
$ swift 1> let a: Int8 = 100 let a: Int8 = 100 2> let b: Int8 = 50 b: Int8 = 50 3> let c: Int8 = 40 c: Int8 = 40 4> (a - c) + b $R0: Int8 = 110 5> (a + b) - c Execution interrupted. Enter code to recover and continue. Enter LLDB commands to investigate (type :help for assistance.)
The famous short-circuit logical operators are really control-flow mechanisms...
Expression | Meaning | Sometimes written as |
---|---|---|
e1 andalso e2
| If e1 if falsy, so is the whole and-expression, so you're done. Otherwise, the result is whatever e2 is. | if e1 then e2 else e1
|
e1 orelse e2
| If e1 if truthy, so is the whole or-expression, so you're done. Otherwise, the result is whatever e2 is. |
if e1 then e1 else e2
|
Short-circuiting appears frequently in many programming idioms
if (p != null && p.key == value) { ... }
function f(x, y) { // Possible in JS, but JS has default arguments so use those! y ??= 1; ... }
open(F, $file) or die "Can't open $file: $!";
exists f && remove f
Note that evaluation order really only matters when side effects can occur (which is why immutability is preferred!). Side effects are what occurs when a storage location is updated or when files or a database are read from or written to.
Storage locations are denoted by lvalues. They are called lvalues because they can appear on the Left side of an assignment. Examples in C:
x
x[4]
y[5].p()->q
*(e1 + e2)
*y[6]
An example of an rvalue is 500 (an integer literal).
Sometimes pretty complicated looking expressions can evaluate to lvalues! Examples:
// C++, but not C (x *= 10) += 7
# Perl ($x < 5 ? $y : $z) = 10;
# Perl $x = "dog"; ${$x} = 2; $x = <STDIN> chomp $x; ${$x} = 2;
(* SML *) val x = ref 0; x := 3; x := !x + 1;
Sometimes, lvalues can be made read-only. Examples:
const
in C, let
in Swift, val
in Kotlin, or final
in Java.
Sometimes, immutability is the default, and you have to add keywords or symbols to make an lvalue mutable.
Initialization and assignment are very different.
One language that makes the distinction explicit in code is C++. Examples:
// C++ Initializations: int x = 10; int y(15); int z(a + 5 / 2); int w(x); Point p(5, 12); Point q = p; // C++ Assignments: x = 12; y = x / 6; q = p; q = midpoint(p1, p2);
Prefer initialization to assignment where possible. Here's a case where you must. Suppose we had a point class in C++:
class Point { public: int x; int y; Point(int x1, int y1): x(x1), y(y1) {} };
Because we defined a constructor with parameters, we cannot ever define uninitialized points:
Point p; // ILLEGAL Point* p = new Point[10]; // ILLEGAL class Rectangle { public: Point corner1, corner2; Rectangle(Point p, Point q) { // ILLEGAL corner1 = p; corner2 = q; } };
The Rectangle constructor failed because it is trying to initialize the fields to their default values and then assign them in the constructor body.... but there is NO default initializer for class Point. You MUST write the Rectangle constructor like this:
class Rectangle { public: Point corner1, corner2; Rectangle(Point p, Point q): corner1(p), corner2(q) {} };
If expressions are only evaluated as-needed, or on demand, or only-if-needed, evaluation is said to be lazy. Otherwise it's eager.
def first(x, y): return x first(f(), g())
Under eager evaluation, both f
and g
are called, and the results of each are passed to first
. Under lazy evaluation, only f
gets called. In our example, suppose f()
evaluates to 3 and g()
to 5. Then:
Eager | Lazy |
---|---|
first(f(), g()) = first(3, f()) = first(3, 5) = 3 |
first(f(), g()) = f() = 3 |
In the cases where g()
crashes or has other side effects, the difference between the two strategies can be a big deal.
A macro is code that gets expanded into new code which then gets compiled and run. In the simplest case, a macro gets expanded into source code, as in this example in C:
#define area(r) (M_PI*(r)*(r)) double f(x) { return 3 / area(x+10); }
but before the program is compiled, the C preprocessor expands the macro, producing:
double f(x) { return 3 / (M_PI*(x+10)*(x+10)); }
#define area(r) M_PI*r*r;Show the expansion of
3/area(x+10)
.
More examples of C macros:
#define MAX(x, y) ((x) > (y) ? (x) : (y)) #define mientras while #define forever while(1)
C macros operate in source code. The macros of Lisp, Clojure, and Julia are much more sophisticated: these operate on abstract syntax trees. We’ll cover these when we get to metaprogramming.
Here are some questions useful for your spaced repetition learning. Many of the answers are not found on this page. Some will have popped up in lecture. Others will require you to do your own research.
-2**2
in JavaScript and Python. Explain why the evaluation produced the value it did in each language.a+(b-c)
and (a+b)-c
produce different results?var x = 5 function f() { x = x * 3 } function g() { x = x * 5 } function h(a, b) { return a + x } print(h(f(), g()))
We’ve covered: