ALGOL 68 is an imperative computer programming language that was conceived as a successor to the ALGOL 60 programming language.
ALGOL 68 (short for ALGOrithmic Language 1968) is an imperative computer programming language that was conceived as a successor to the ALGOL 60 programming language, designed with the goal of a much wider scope of application and more rigorously defined syntax and semantics.
The "yield" statement in python allows simple iteration from a procedure, and it also means that sequences don't need to be pre-calculated AND stored in a array of "arbitrary" size.
Is there a there a similar way of iterating (with yield) from a C…
I'm doing this simple "game engine" for cli pixel games creation, a68g game engine.
I have seen that in the lib/canvas.a68 file. Those 2 loops are making a error:
7 arr := new
1
a68g: lib/operators.a68: runtime error: 1: insufficient…
I'm re-visiting Algol68 because I think it's historically important. I would
like to link one of my programs (written as a student) to the NAG library routine f07abf
(linear equation solver). The question is, being a retired academic, do I need to…
I have 2 flavors of the same ALGOL code - its a ONE-ONE replacement
Which uses - RESIZE (TO RETURN IT LIBARAY POOL)
Which uses - DEALLOCATE (TO RETURN IT SYSTEM)
The one which the DEALLOCATE Consumes more CPU time and inturn more %processor…
Here is a relatively simple code for "Evaluation of pi using the Mid-ordinate Rule on a quadrant of circle with radius 2 units."
main.alg
BEGIN
REAL x, y, sumy, pi;
INT n := lowerlimit, p := 1, lowerlimit := 10, upperlimit := 100, interval :=…
Suppose we have the following instructions in Algol 68 and C, respectively:
ref int x := 5; and (int*) x = 5;. What are their semantic difference?, it's the same?, because I think that the second one says something like "x will point to a constant"…
Suppose I have an algol-like language, with static types and the following piece of code:
a := b + c * d;
where a is a float, b an integer, c a double and d a long. Then, the language will convert d to long to operate with c, and b to double to…
I'm studying semaphores. In the book there's a question without an answer, and I would really like to know how to solve the problem.
The question is:
This is a counting semaphore written in Algol68:
Down mutex: if mutex = 0 then access is blocked
…
I am suppose the change the code below to do factorial. Anyone can help me with it?
This is the Fibonacci.
PROC print fibo = (INT n) VOID:
BEGIN
INT a:= 0, b = 1;
FOR i FROM 1 TO n DO
print((whole(i,0), "==>", whole(b,0), new…
Curly Bracket languages are well known: (wikipedia)
Other programming languages can have BEGIN ~ END vs LIVE ~ EVIL block structuring. eg
A) BEGIN ~ END, DO ~ END, IF ~ END IF - examples:
Ada, Modula, Pascal, PL/I, Ruby
etc...
B) IF ~ FI, DO…
Suggest me on how to change the following code into c++ code:
ROOTPROC VarUse
PROC VarUse
ROOT Cfile;
1 {
2 [
3 (?NameRef
4 (IF (AND (HAS-TYPE $parent Assignment) (IS-EQUAL $slot ``lhs''))
5 (THEN (PRINT stdout "Variable %s defined at %s" $token…
like in Algol 68 an if statement ends with fi
and a case statement ends with esac , so what are some readability problem other than the reserved words doesn't really make sense in English.
Any suggestions would be appreciated
