Prolog Operator =:=

Question

There are some special operators in Prolog, one of them is "is", however, recently I came across the =:= operators, and I have no idea how it works.

Can someone explain what the operator does, and also where can I find a predefined list of such special operators and what they do?

Thanks.

Answer 1

?- 2+3 =:= 6-1.
true.

?- 2+3 is 6-1.
false.

Also please see docs http://www.swi-prolog.org/pldoc/man?predicate=is/2

Answer 2

I think the above answer deserves a few words of explanation here nevertheless.

A short note in advance: Arithmetic expressions in Prolog are just terms ("Everything is a term in Prolog"), which are not evaluated automatically. (If you have a Lisp background, think of quoted lists). So 3 + 4 is just the same as +(3,4), which does nothing on its own. It is the responsibility of individual predicates to evaluate those terms.

Several built-in predicates do implicit evaluation, among them the arithmetic comparsion operators like =:= and is. While =:= evaluates both arguments and compares the result, is accepts and evaluates only its right argument as an arithmetic expression.

The left argument has to be an atom, either a numeric constant (which is then compared to the result of the evaluation of the right operand), or a variable. If it is a bound variable, its value has to be numeric and is compared to the right operand as in the former case. If it is an unbound variable, the result of the evaluation of the right operand is bound to that variable. is is often used in this latter case, to bind variables.

To pick up on an example from the above linked Prolog Dictionary: To test if a number N is even, you could use both operators:

0 is N mod 2  % true if N is even
0 =:= N mod 2 % dito

But if you want to capture the result of the operation you can only use the first variant. If X is unbound, then:

X is N mod 2   % X will be 0 if N is even
X =:= N mod 2  % !will bomb with argument/instantiation error!

Rule of thumb: If you just need arithmetic comparison, use =:=. If you want to capture the result of an evaluation, use is.

Answer 3

Complementing the existing answers, I would like to state a few additional points:

An operator is an operator

First of all, the operator =:= is, as the name indicates, an operator. In Prolog, we can use the predicate current_op/3 to learn more about operators. For example:

?- current_op(Prec, Type, =:=).
Prec = 700,
Type = xfx.

This means that the operator =:= has precedence 700 and is of type xfx. This means that it is a binary infix operator.

This means that you can, if you want, write a term like =:=(X, Y) equivalently as X =:= Y. In both cases, the functor of the term is =:=, and the arity of the term is 2. You can use write_canonical/1 to verify this:

?- write_canonical(a =:= b).
=:=(a,b)

A predicate is not an operator

So far, so good! This has all been a purely syntactical feature. However, what you are actually asking about is the predicate (=:=)/2, whose name is =:= and which takes 2 arguments.

As others have already explained, the predicate (=:=)/2 denotes arithmetic equality of two arithmetic expressions. It is true iff its arguments evaluate to the same number.

For example, let us try the most general query, by which we ask for any solution whatsoever, using variables as arguments:

?- X =:= Y.
ERROR: Arguments are not sufficiently instantiated

Hence, this predicate is not a true relation, since we cannot use it for generating results! This is a quite severe drawback of this predicate, clashing with what you commonly call "declarative programming".

The predicate only works in the very specific situation that both arguments are fully instantiated. For example:

?- 1 + 2 =:= 3.
true.

We call such predicates moded because they can only be used in particular modes of usage. For the vast majority of beginners, moded predicates are a nightmare to use, because they require you to think about your programs procedurally, which is quite hard at first and remains hard also later. Also, moded predicates severely limit the generality of your programs, because you cannot use them on all directions in which you could use pure predicates.

Constraints are a more general alternative

Prolog also provides much more general arithmetic predicates in the form of arithmetic constraints.

For example, in the case of integers, try your Prolog system's CLP(FD) constraints. One of the most important CLP(FD) constraints denotes arithmetic equality and is called (#=)/2. In complete analogy to (=:=)/2, the operator (#=)/2 is also defined as an infix operator, and so you can write for example:

| ?- 1 + 2 #= 3.

yes

I am using GNU Prolog as one particular example, and many other Prolog systems also provide CLP(FD) implementations.

A major attraction of constraints is found in their generality. For example, in contrast to (=:=)/2, we get with the predicate (#=)/2:

| ?- X + 2 #= 3.

X = 1

| ?- 1 + Y #= 3.

Y = 2

And we can even ask the most general query:

| ?- X #= Y.

X = _#0(0..268435455)
Y = _#0(0..268435455)

Note how naturally these predicates blend into Prolog and act as relations between integer expressions that can be queried in all directions.

Depending on the domain of interest, my recommendition is to use CLP(FD), CLP(Q), CLP(B) etc. instead of using more low-level arithmetic predicates.

Also see clpfd, clpq and clpb for more information.

Coincidentally, the operator =:= is used by CLP(B) with a completely different meaning:

?- sat(A =:= B+1).
A = 1,
sat(B=:=B).

This shows that you must distinguish between operators and predicates. In the above case, the predicate sat/1 has interpreted the given expression as a propositional formula, and in this context, =:= denotes equality of Boolean expressions.

Answer 4

I found my own answer, http://www.cse.unsw.edu.au/~billw/prologdict.html

Answer 5

Its an ISO core standard predicate operator, which cannot be bootstrapped from unification (=)/2 or syntactic equality (==)/2. It is defined in section 8.7 Arithmetic Comparison. And it basically behaves as follows:

E =:= F :- 
    X is E, 
    Y is F, 
    arithmetic_compare(=, X, Y).

So both the left hand side (LHS) and right hand side (RHS) must be arithmetic expressions that are evaluted before they are compared. Arithmetic comparison can compare across numeric types. So we have:

   GNU Prolog 1.4.5 (64 bits)

   ?- 0 = 0.0.
   no

   ?- 0 == 0.0
   no

   ?- 0 =:= 0.0.
   yes

Answer 6

=:= is a comparison operator.A1 =:= A2 succeeds if values of expressions A1 and A2 are equal. A1 == A2 succeeds if terms A1 and A2 are identical;

Answer 7

First operator =:= is check equal ? for example enter image description here

it"s return true. but this returns false enter image description here