Priority queue: Difference between revisions

=={{header|Phix}}==

Dictionary based solution. Allows duplicate tasks, FIFO within priority, and uses a callback-style method of performing tasks.<br>

Assumes 5 is the highest priority and should be done first, for 1 first just delete the ",true" on traverse_dict calls.

<lang Phix>integer tasklist = new_dict()

procedure add_task(integer priority, string desc)

integer k = getd_index(priority,tasklist)

if k=0 then

putd(priority,{desc},tasklist)

else

sequence descs = getd_by_index(k,tasklist)

putd(priority,append(descs,desc),tasklist)

end if

end procedure

function list_task_visitor(integer priority, sequence descs, integer /*user_data*/)

?{priority,descs}

return 1

end function

procedure list_tasks()

traverse_dict(routine_id("list_task_visitor"), 0, tasklist,true)

end procedure

function pop_task_visitor(integer priority, sequence descs, integer rid)

string desc = descs[1]

descs = descs[2..$]

if length(descs)=0 then

deld(priority,tasklist)

else

putd(priority,descs,tasklist)

end if

call_proc(rid,{priority,desc})

return 0

end function

procedure pop_task(integer rid)

if dict_size(tasklist)!=0 then

traverse_dict(routine_id("pop_task_visitor"), rid, tasklist,true)

end if

end procedure

add_task(3,"Clear drains")

add_task(4,"Feed cat")

add_task(5,"Make tea")

add_task(1,"Solve RC tasks")

add_task(2,"Tax return")

procedure do_task(integer priority, string desc)

?{priority,desc}

end procedure

list_tasks()

?"==="

pop_task(routine_id("do_task"))

?"==="

list_tasks()</lang>

{{out}}

<pre>

{5,{"Make tea"}}

{4,{"Feed cat"}}

{3,{"Clear drains"}}

{2,{"Tax return"}}

{1,{"Solve RC tasks"}}

"==="

{5,"Make tea"}

"==="

{4,{"Feed cat"}}

{3,{"Clear drains"}}

{2,{"Tax return"}}

{1,{"Solve RC tasks"}}

</pre>