Related
The problem
Consider the following script written in Maxima
$ cat main.max
foo(a) ::= block(
if a = 0 then return(0) else return(1)
)$
bar(a) ::= block(
return(foo(a))
)$
foo(0);
bar(0);
Executing this script yields to
$ cat main.max | maxima --very-quiet
0
1
The question
Isn't it supposed that calling foo by itself results the same as if it is called from another function (in this scenario, bar)?
Here there is another example
$ cat main.max
foo(a) ::= block(disp(a))$
bar(a) ::= block(foo(a))$
foo(0)$
bar(0)$
$ cat main.max | maxima --very-quiet
0
a
In other words: Why isn't maxima replacing the argument that is passed to bar in the foo function call which is located within bar?
Additional context
This is the version that I'm currently using
$ maxima --version
Maxima 5.43.2
::= defines a function which quotes (does not evaluate) its arguments, and the return value of the function is evaluated by the caller. Such a function is conventionally called a "macro" for historical reasons.
I think you want an ordinary, argument-evaluating function, which is defined by :=. Try substituting := for ::= in your function definitions -- when I try that, I get 0 for bar(0) as expected.
Macros are useful, but in relatively narrow circumstances. I think ordinary functions are much more common than macros.
As an aside, in the functions which you have shown, block and return are unneeded. My advice is to just leave them out and make the functions more succinct and therefore more clear. E.g.
foo(a) := if a = 0 then 0 else 1;
bar(a) := foo(a);
Finally, note that = is only literal comparison, not equivalence; equivalence is tested with equal. E.g. suppose x is not otherwise defined. Then is(x = 0) yields false, but is(equal(x, 0)) yields the expression equal(x, 0). Depending on what you're doing, one or the other might be appropriate.
In Python, what is the difference between expressions and statements?
Expressions only contain identifiers, literals and operators, where operators include arithmetic and boolean operators, the function call operator () the subscription operator [] and similar, and can be reduced to some kind of "value", which can be any Python object. Examples:
3 + 5
map(lambda x: x*x, range(10))
[a.x for a in some_iterable]
yield 7
Statements (see 1, 2), on the other hand, are everything that can make up a line (or several lines) of Python code. Note that expressions are statements as well. Examples:
# all the above expressions
print 42
if x: do_y()
return
a = 7
Expression -- from the New Oxford American Dictionary:
expression: Mathematics a collection
of symbols that jointly express a
quantity : the expression for the
circumference of a circle is 2πr.
In gross general terms: Expressions produce at least one value.
In Python, expressions are covered extensively in the Python Language Reference In general, expressions in Python are composed of a syntactically legal combination of Atoms, Primaries and Operators.
Python expressions from Wikipedia
Examples of expressions:
Literals and syntactically correct combinations with Operators and built-in functions or the call of a user-written functions:
>>> 23
23
>>> 23l
23L
>>> range(4)
[0, 1, 2, 3]
>>> 2L*bin(2)
'0b100b10'
>>> def func(a): # Statement, just part of the example...
... return a*a # Statement...
...
>>> func(3)*4
36
>>> func(5) is func(a=5)
True
Statement from Wikipedia:
In computer programming a statement
can be thought of as the smallest
standalone element of an imperative
programming language. A program is
formed by a sequence of one or more
statements. A statement will have
internal components (e.g.,
expressions).
Python statements from Wikipedia
In gross general terms: Statements Do Something and are often composed of expressions (or other statements)
The Python Language Reference covers Simple Statements and Compound Statements extensively.
The distinction of "Statements do something" and "expressions produce a value" distinction can become blurry however:
List Comprehensions are considered "Expressions" but they have looping constructs and therfore also Do Something.
The if is usually a statement, such as if x<0: x=0 but you can also have a conditional expression like x=0 if x<0 else 1 that are expressions. In other languages, like C, this form is called an operator like this x=x<0?0:1;
You can write you own Expressions by writing a function. def func(a): return a*a is an expression when used but made up of statements when defined.
An expression that returns None is a procedure in Python: def proc(): pass Syntactically, you can use proc() as an expression, but that is probably a bug...
Python is a bit more strict than say C is on the differences between an Expression and Statement. In C, any expression is a legal statement. You can have func(x=2); Is that an Expression or Statement? (Answer: Expression used as a Statement with a side-effect.) The assignment statement of x=2 inside of the function call of func(x=2) in Python sets the named argument a to 2 only in the call to func and is more limited than the C example.
Though this isn't related to Python:
An expression evaluates to a value.
A statement does something.
>>> x + 2 # an expression
>>> x = 1 # a statement
>>> y = x + 1 # a statement
>>> print y # a statement (in 2.x)
2
An expression is something that can be reduced to a value, for example "1+3" is an expression, but "foo = 1+3" is not.
It's easy to check:
print(foo = 1+3)
If it doesn't work, it's a statement, if it does, it's an expression.
Another statement could be:
class Foo(Bar): pass
as it cannot be reduced to a value.
Statements represent an action or command e.g print statements, assignment statements.
print 'hello', x = 1
Expression is a combination of variables, operations and values that yields a result value.
5 * 5 # yields 25
Lastly, expression statements
print 5*5
An expression is something, while a statement does something.
An expression is a statement as well, but it must have a return.
>>> 2 * 2 #expression
>>> print(2 * 2) #statement
PS:The interpreter always prints out the values of all expressions.
An expression is a statement that returns a value. So if it can appear on the right side of an assignment, or as a parameter to a method call, it is an expression.
Some code can be both an expression or a statement, depending on the context. The language may have a means to differentiate between the two when they are ambiguous.
STATEMENT:
A Statement is a action or a command that does something. Ex: If-Else,Loops..etc
val a: Int = 5
If(a>5) print("Hey!") else print("Hi!")
EXPRESSION:
A Expression is a combination of values, operators and literals which yields something.
val a: Int = 5 + 5 #yields 10
Expressions always evaluate to a value, statements don't.
e.g.
variable declaration and assignment are statements because they do not return a value
const list = [1,2,3];
Here we have two operands - a variable 'sum' on the left and an expression on the right.
The whole thing is a statement, but the bit on the right is an expression as that piece of code returns a value.
const sum = list.reduce((a, b)=> a+ b, 0);
Function calls, arithmetic and boolean operations are good examples of expressions.
Expressions are often part of a statement.
The distinction between the two is often required to indicate whether we require a pice of code to return a value.
References
Expressions and statements
2.3 Expressions and statements - thinkpython2 by Allen B. Downey
2.10. Statements and Expressions - How to Think like a Computer Scientist by Paul Resnick and Brad Miller
An expression is a combination of values, variables, and operators. A value all by itself is
considered an expression, and so is a variable, so the following are all legal expressions:
>>> 42
42
>>> n
17
>>> n + 25
42
When you type an expression at the prompt, the interpreter evaluates it, which means that
it finds the value of the expression. In this example, n has the value 17 and n + 25 has the
value 42.
A statement is a unit of code that has an effect, like creating a variable or displaying a
value.
>>> n = 17
>>> print(n)
The first line is an assignment statement that gives a value to n. The second line is a print
statement that displays the value of n.
When you type a statement, the interpreter executes it, which means that it does whatever
the statement says. In general, statements don’t have values.
An expression translates to a value.
A statement consumes a value* to produce a result**.
*That includes an empty value, like: print() or pop().
**This result can be any action that changes something; e.g. changes the memory ( x = 1) or changes something on the screen ( print("x") ).
A few notes:
Since a statement can return a result, it can be part of an expression.
An expression can be part of another expression.
Statements before could change the state of our Python program: create or update variables, define function, etc.
And expressions just return some value can't change the global state or local state in a function.
But now we got :=, it's an alien!
Expressions:
Expressions are formed by combining objects and operators.
An expression has a value, which has a type.
Syntax for a simple expression:<object><operator><object>
2.0 + 3 is an expression which evaluates to 5.0 and has a type float associated with it.
Statements
Statements are composed of expression(s). It can span multiple lines.
A statement contains a keyword.
An expression does not contain a keyword.
print "hello" is statement, because print is a keyword.
"hello" is an expression, but list compression is against this.
The following is an expression statement, and it is true without list comprehension:
(x*2 for x in range(10))
Python calls expressions "expression statements", so the question is perhaps not fully formed.
A statement consists of pretty much anything you can do in Python: calculating a value, assigning a value, deleting a variable, printing a value, returning from a function, raising an exception, etc. The full list is here: http://docs.python.org/reference/simple_stmts.html#
An expression statement is limited to calling functions (e.g.,
math.cos(theta)"), operators ( e.g., "2+3"), etc. to produce a value.
I would like to know, how expressions are parsed when are mixed with control flow.
Let's assume such syntax:
case
when a == Method() + 1
then Something(1)
when a == Other() - 2
then 1
else 0
end
We've got here two conditional expressions, Method() + 1, Something(1) and 0. Each can be translated to postfix by Shunting-yard algorithm and then easily translated into AST. But is it possible to extend this algorithm to handle control - flow also? Or are there other approaches to solve such mixing of expressions and control flows?
another example:
a == b ? 1 : 2
also how can I classify such expression: a between b and c, can I say that between is three arguments function? Or is there any special name for such expressions?
You can certainly parse the ternary operator with an operator-precedence grammar. In
expr ? expr : expr
the binary "operator" here is ? expr :, which conveniently starts and ends with an operator token (albeit different ones). To adapt shunting yard to that, assign the right-precedence of ? and the left-precedence of : to the precedence of the ?: operator. The left-precedence of ? and the right-precedence of : are ±∞, just like parentheses (which, in effect, they are).
Since the case statement is basically repeated application of the ternary operator, using slightly different spellings for the tokens, and yields to a similar solution. (Here case when and end are purely parenthetic, while then and the remaining whens correspond to ? and :.)
Having said that, it really is simpler to use an LALR(1) parser generator, and there is almost certainly one available for whatever language you are writing in.
It's clear that both the ternary operator and OP's case statement are operator grammars:
Ternary operator:
ternary-expr: non-ternary-expr
| non-ternary-expr '?' expr ':' ternary-expr
Normally, the ternary operator will be lower precedence from any other operator and associate to the right, which is how the above is written. In C and other languages ternary expressions have the same precedence as assignment expressions, which is straightforward to add. That results in the relationships
X ·> ?
? <· X
? ·=· :
X ·> :
: <· X
Case statement (one of many possible formulations):
case_statement: 'case' case_body 'else' expr 'end'
case_body: 'when' expr 'then' expr
| case_body 'when' expr 'then' expr
Here are the precedence relationships for the above grammar:
case <· when
case ·=· else
when <· X (see below)
when ·=· then
then ·> when
then ·> else
else <· X
else ·=· end
X ·> then
X ·> when
X ·> end
X in the above relations refers to any binary or unary operator, any value terminal, ( and ).
It's straightforward to find left- and right-precedence functions for all of those terminals; the pattern will be similar to that of parentheses in a standard algebraic grammar.
The Shunting-yard algorithm is for expressions with unary and binary operators. You need something more powerful such as LL(1) or LALR(1) to parse control flow statements, and once you have that it will also handle expressions as well. No need for the Shunting-yard algorithm at all.
Is there any algorithm that can find the sign of an arbitrary symbolic algebraic expression given in a "Tree - Form"?
I know that a general algorithm doesn't exist because the zero recognizion problem is undecidable for an arbitrary expression, but how should I approach the problem of finding the sign of an expression?
(how is this done in computer algebra?)
For example: sign(sqrt(2)-1) = ?
Evaluate the function value
You need function evaluator engine for that (it is not that hard to code) there is no way to evaluate sign only if you want to support +,- operations !!! All my function evaluators works like this:
compile the source text of the function
First create supported functions table (id,num of operands,name,pointer to function) like:
+,-,*,/,sin,cos,....
These will be the building blocks to any supported expression you need to evaluate. Do not forget to code all functions in your code too. Handle brackets (,) also as functions (push,pop). Group your functions by number of operands so +,- are with 1 and 2 operands (two different functions each !!!).
Now from expression extract:
variable names
constant names and values
number values
Into some kind of table/list:
variables[](id,name,value)
constants[](id,name,value)
numbers [](id, ,value)
And now finally construct compiled function string. My strings are set of two int's. First is type (which table to use) and second is id (index in table).
for example expression:
sign(sqrt(2)-1)
types:
id type
0 function
1 number
2 constant
3 variable
functions:
id name pointer
0 '(' ???
1 ')' ???
2 '+' ???
3 '-' ???
4 '*' ???
5 '/' ???
6 'sqrt' ???
7 'sign' ???
There are no variables or constants. The numbers are:
id value
0 2
1 1
compiled string:
type id
0 7 // sign(1 operand)
0 6 // sqrt(1 operand)
1 0 // 2
0 3 // - (2 operands)
1 1 // 1
After compilation you need to interpret the string and evaluate it's value.
init variables
op1=0`,`op2=0, // set all operands to zero (number depends on supported functions usually 2)
opn=0 // actual operands number
fx=none // actual function (for example none=-1)
fxn=0 // actual function operands number
read first record of compiled string
if it is value (number,constant,variable) set appropriate op? value with it and increment operand counter opn++.
if it is function set fx,fxn code with it
if opn == fxn
You reached needed operand count so execute function fx and init next function
op1=fxtab[fx].pointer(op1,op2,...)
fx=none,fxn=1
opn=1 (some spec functions can return more operands, then set op1,op2,.. opn=...)
if not end of string goto #2 but with next string record
at the end op1 should hold your output value
some example functions (C++ implementation):
double sign(double op1)
{
if (op1>0.0) return +1.0;
if (op1<0.0) return -1.0;
return 0.0;
}
double sqrt1(double op1) { return sqrt(op1); }
double plus1(double op1) { return op1; }
double minus1(double op1) { return -op1; }
double plus2(double op1,double op2) { return op1+op2; }
double minus2(double op1,double op2) { return op1-op2; }
[Notes]
You have to handle special cases like function = "";. Also beware spacing, case sensitivity because any error in compilation invalidates the result.
Speed is not a big issue this is interpreting-evaluation not numerical solution. All operations are called the same times as you would do on the paper.
You should also handle mathematic errors (overflows,invalid operands,NaN,Inf ...)
I usually group functions with the same number of operands to own type to simplify things
Where can I find a list of Scala's "magic" functions, such as apply, unapply, update, +=, etc.?
By magic-functions I mean functions which are used by some syntactic sugar of the compiler, for example
o.update(x,y) <=> o(x) = y
I googled for some combination of scala magic and synonyms of functions, but I didn't find anything.
I'm not interested with the usage of magic functions in the standard library, but in which magic functions exists.
As far as I know:
Getters/setters related:
apply
update
identifier_=
Pattern matching:
unapply
unapplySeq
For-comprehensions:
map
flatMap
filter
withFilter
foreach
Prefixed operators:
unary_+
unary_-
unary_!
unary_~
Beyond that, any implicit from A to B. Scala will also convert A <op>= B into A = A <op> B, if the former operator isn't defined, "op" is not alphanumeric, and <op>= isn't !=, ==, <= or >=.
And I don't believe there's any single place where all of Scala's syntactic sugars are listed.
In addition to update and apply, there are also a number of unary operators which (I believe) qualify as magical:
unary_+
unary_-
unary_!
unary_~
Add to that the regular infix/suffix operators (which can be almost anything) and you've got yourself the complete package.
You really should take a look at the Scala Language Specification. It is the only authoritative source on this stuff. It's not that hard to read (as long as you're comfortable with context-free grammars), and very easily searchable. The only thing it doesn't specify well is the XML support.
Sorry if it's not exactly answering your question, but my favorite WTF moment so far is # as assignment operator inside pattern match. Thanks to soft copy of "Programming in Scala" I found out what it was pretty quickly.
Using # we can bind any part of a pattern to a variable, and if the pattern match succeeds, the variable will capture the value of the sub-pattern. Here's the example from Programming in Scala (Section 15.2 - Variable Binding):
expr match {
case UnOp("abs", e # UnOp("abs", _)) => e
case _ =>
}
If the entire pattern match succeeds,
then the portion that matched the
UnOp("abs", _) part is made available
as variable e.
And here's what Programming Scala says about it.
That link no longer works. Here is one that does.
I'll also add _* for pattern matching on an arbitrary number of parameters like
case x: A(_*)
And operator associativity rule, from Odersky-Spoon-Venners book:
The associativity of an operator in Scala is determined by its last
character. As mentioned on <...>, any method that ends
in a ‘:’ character is invoked on its right operand, passing in the
left operand. Methods that end in any other character are the other
way around. They are invoked on their left operand, passing in the
right operand. So a * b yields a.*(b), but a ::: b yields b.:::(a).
Maybe we should also mention syntactic desugaring of for expressions which can be found here
And (of course!), alternative syntax for pairs
a -> b //converted to (a, b), where a and b are instances
(as correctly pointed out, this one is just an implicit conversion done through a library, so it's probably not eligible, but I find it's a common puzzler for newcomers)
I'd like to add that there is also a "magic" trait - scala.Dynamic:
A marker trait that enables dynamic invocations. Instances x of this trait allow method invocations x.meth(args) for arbitrary method names meth and argument lists args as well as field accesses x.field for arbitrary field names field.
If a call is not natively supported by x (i.e. if type checking fails), it is rewritten according to the following rules:
foo.method("blah") ~~> foo.applyDynamic("method")("blah")
foo.method(x = "blah") ~~> foo.applyDynamicNamed("method")(("x", "blah"))
foo.method(x = 1, 2) ~~> foo.applyDynamicNamed("method")(("x", 1), ("", 2))
foo.field ~~> foo.selectDynamic("field")
foo.varia = 10 ~~> foo.updateDynamic("varia")(10)
foo.arr(10) = 13 ~~> foo.selectDynamic("arr").update(10, 13)
foo.arr(10) ~~> foo.applyDynamic("arr")(10)
As of Scala 2.10, defining direct or indirect subclasses of this trait is only possible if the language feature dynamics is enabled.
So you can do stuff like
import scala.language.dynamics
object Dyn extends Dynamic {
def applyDynamic(name: String)(a1: Int, a2: String) {
println("Invoked " + name + " on (" + a1 + "," + a2 + ")");
}
}
Dyn.foo(3, "x");
Dyn.bar(3, "y");
They are defined in the Scala Language Specification.
As far as I know, there are just three "magic" functions as you mentioned.
Scalas Getter and Setter may also relate to your "magic":
scala> class Magic {
| private var x :Int = _
| override def toString = "Magic(%d)".format(x)
| def member = x
| def member_=(m :Int){ x = m }
| }
defined class Magic
scala> val m = new Magic
m: Magic = Magic(0)
scala> m.member
res14: Int = 0
scala> m.member = 100
scala> m
res15: Magic = Magic(100)
scala> m.member += 99
scala> m
res17: Magic = Magic(199)