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[英]Why does Haskell perform so poorly when executing C-like codes? (in this instance at least)
[英]Does Prolog have an alias “operator” like Haskell?
在Haskell中,有一種稱為“as”-operator的語言特性(有時稱為別名)。 這個想法如下:假設您有一個函數,例如將列表作為輸入並希望返回所有尾部,您可以將其實現為:
tails a@(_:xs) = a : tails xs
tails [] = [[]]
@
確保您既可以引用整個參數,也可以引用參數結構的某些部分。 這是智能性能(它更像是性能破解,因為在第一行的主體中重構數組(x:xs)
),如果沒有被編譯器優化,將導致分配新對象,修改字段等。有關更多信息,請參見此處 。
我想知道Prolog是否有相同的東西:例如,如果你想在Prolog中實現尾部,可以通過以下方式完成:
tails([H|T],[[H|T]|TA]) :-
tails(T,TA).
tails([],[[]]).
但如果有一個“as”運算符,它可能會更有效:
tails(L@[_|T],[L|TA]) :- %This does not compile
tails(T,TA).
tails([],[[]]).
有沒有這樣的結構或語言擴展?
TL; DR:好主意1 ! 加速似乎限制在約20%(對於大多數列表大小)。
在這個答案中,我們比較了三個不同的謂詞,它們在@
-like數據重用方面有所不同:
list_tails([], [[]]). % (1) like `tails/2` given by the OP ... list_tails([E|Es], [[E|Es]|Ess]) :- % ....... but with a better name :-) list_tails(Es, Ess). list_sfxs1(Es, [Es|Ess]) :- % (2) "re-use, mutual recursion" aux_list_sfxs1(Es, Ess). % "sfxs" is short for "suffixes" aux_list_sfxs1([], []). aux_list_sfxs1([_|Es], Ess) :- list_sfxs1(Es, Ess). list_sfxs2([], [[]]). % (3) "re-use, direct recursion" list_sfxs2(Es0, [Es0|Ess]) :- Es0 = [_|Es], list_sfxs2(Es, Ess).
要測量運行時,我們使用以下代碼:
:-( dif(D,sicstus), current_prolog_flag(dialect,D) ; use_module(library(between)) ). run_benchs(P_2s, P_2, L, N, T_ms) :- between(1, 6, I), L is 10^I, N is 10^(8-I), length(Xs, L), member(P_2, P_2s), garbage_collect, call_walltime(run_bench_core(P_2,Xs,N), T_ms). run_bench_core(P_2, Xs, N) :- between(1, N, _), call(P_2, Xs, _), false. run_bench_core(_, _, _).
為了測量walltime 2,我們利用call_ walltime /2
2的變異-a call_time/2
:
call_walltime(G, T_ms) :- statistics(walltime, [T0|_]), G, statistics(walltime, [T1|_]), T_ms is T1 - T0.
讓我們將上面的代碼變體放到測試中......
L
... N
次(為了更好的准確性)。 首先,我們使用swi-prolog版本7.3.14(64位):
?- run_benchs([list_sfxs1,list_sfxs2,list_tails], P_2, L, N, T_ms). P_2 = list_sfxs1, L*N = 10*10000000, T_ms = 7925 ; P_2 = list_sfxs2, L*N = 10*10000000, T_ms = 7524 ; P_2 = list_tails, L*N = 10*10000000, T_ms = 6936 ; P_2 = list_sfxs1, L*N = 100*1000000, T_ms = 6502 ; P_2 = list_sfxs2, L*N = 100*1000000, T_ms = 5861 ; P_2 = list_tails, L*N = 100*1000000, T_ms = 5618 ; P_2 = list_sfxs1, L*N = 1000*100000, T_ms = 6434 ; P_2 = list_sfxs2, L*N = 1000*100000, T_ms = 5817 ; P_2 = list_tails, L*N = 1000*100000, T_ms = 9916 ; P_2 = list_sfxs1, L*N = 10000*10000, T_ms = 6328 ; P_2 = list_sfxs2, L*N = 10000*10000, T_ms = 5688 ; P_2 = list_tails, L*N = 10000*10000, T_ms = 9442 ; P_2 = list_sfxs1, L*N = 100000*1000, T_ms = 10255 ; P_2 = list_sfxs2, L*N = 100000*1000, T_ms = 10296 ; P_2 = list_tails, L*N = 100000*1000, T_ms = 14592 ; P_2 = list_sfxs1, L*N = 1000000*100, T_ms = 6955 ; P_2 = list_sfxs2, L*N = 1000000*100, T_ms = 6534 ; P_2 = list_tails, L*N = 1000000*100, T_ms = 9738.
然后,我們使用sicstus-prolog版本4.3.2(64位)重復上一個查詢3 :
?- run_benchs([list_sfxs1,list_sfxs2,list_tails], P_2, L, N, T_ms). P_2 = list_sfxs1, L*N = 10*10000000, T_ms = 1580 ; P_2 = list_sfxs2, L*N = 10*10000000, T_ms = 1610 ; P_2 = list_tails, L*N = 10*10000000, T_ms = 1580 ; P_2 = list_sfxs1, L*N = 100*1000000, T_ms = 710 ; P_2 = list_sfxs2, L*N = 100*1000000, T_ms = 750 ; P_2 = list_tails, L*N = 100*1000000, T_ms = 840 ; P_2 = list_sfxs1, L*N = 1000*100000, T_ms = 650 ; P_2 = list_sfxs2, L*N = 1000*100000, T_ms = 660 ; P_2 = list_tails, L*N = 1000*100000, T_ms = 740 ; P_2 = list_sfxs1, L*N = 10000*10000, T_ms = 620 ; P_2 = list_sfxs2, L*N = 10000*10000, T_ms = 650 ; P_2 = list_tails, L*N = 10000*10000, T_ms = 740 ; P_2 = list_sfxs1, L*N = 100000*1000, T_ms = 670 ; P_2 = list_sfxs2, L*N = 100000*1000, T_ms = 650 ; P_2 = list_tails, L*N = 100000*1000, T_ms = 750 ; P_2 = list_sfxs1, L*N = 1000000*100, T_ms = 12610 ; P_2 = list_sfxs2, L*N = 1000000*100, T_ms = 12560 ; P_2 = list_tails, L*N = 1000000*100, T_ms = 33460.
摘要:
腳注1:為什么把(@)/2
放在規則頭上的噱頭? 最終得到非慣用的 Prolog代碼?
腳注2:我們對總運行時間感興趣。 為什么? 因為垃圾收集成本顯示更大的數據量!
腳注3:為了便於閱讀,答案序列已經過后處理。
腳注4:自4.3.0版以來可用。 目前的目標架構包括IA-32和AMD64 。
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