blocks|key|2026521|text|out+of+local+stack错误意味着程序使用了太多的内存并超出了分配的空间；当程序陷入无限循环时，经常会发生这种情况。在您的示例中，跟踪如下：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2026522|[trace]+2+?-+fib(2,M).
+++Call:+(6)+fib(2,+_G671)+?+creep
%5E++Call:+(7)+_G746+is+2%2B+-1+?+creep
%5E++Exit:+(7)+1+is+2%2B+-1+?+creep
%5E++Call:+(7)+_G749+is+2%2B+-2+?+creep
%5E++Exit:+(7)+0+is+2%2B+-2+?+creep
+++Call:+(7)+fib(1,+_G747)+?+creep
+++Exit:+(7)+fib(1,+1)+?+creep
+++Call:+(7)+fib(0,+_G747)+?+creep
+++Exit:+(7)+fib(0,+0)+?+creep
%5E++Call:+(7)+_G671+is+1%2B0+?+creep
%5E++Exit:+(7)+1+is+1%2B0+?+creep
+++Exit:+(6)+fib(2,+1)+?+creep
M+=+1+;
+++Redo:+(7)+fib(0,+_G747)+?+creep
%5E++Call:+(8)+_G752+is+0%2B+-1+?+creep
%5E++Exit:+(8)+-1+is+0%2B+-1+?+creep
%5E++Call:+(8)+_G755+is+0%2B+-2+?+creep
%5E++Exit:+(8)+-2+is+0%2B+-2+?+creep
+++Call:+(8)+fib(-1,+_G753)+?+creep
%5E++Call:+(9)+_G758+is+-1%2B+-1+?+creep
%5E++Exit:+(9)+-2+is+-1%2B+-1+?+creep
%5E++Call:+(9)+_G761+is+-1%2B+-2+?+creep
%5E++Exit:+(9)+-3+is+-1%2B+-2+?+creep
+++Call:+(9)+fib(-2,+_G759)+?+creep
%5E++Call:+(10)+_G764+is+-2%2B+-1+?+creep
%5E++Exit:+(10)+-3+is+-2%2B+-1+?+creep
...|code-block|syntax|javascript|2026523|正如你所看到的，在发现第二个斐波纳契数是1(根据你的定义)之后，你要求第二个解；因为你没有指定第三个子句只能在N>1+prolog试图通过计算fib(-1)，fib(-2)，fib(-3)等找到第二个解时使用。|2026524|要解决这个问题，必须在第三个子句中添加N>1或类似的规则|2026525|entityMap^0|0|I|0|0|0|J|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

The <code>out of local stack</code> error means that the program used too much memory and exceeded the allocated space; this occurs often when the program falls in an infinite loop. In your case, here is the trace:

<pre><code>[trace] 2 ?- fib(2,M).
 Call: (6) fib(2, _G671) ? creep
^ Call: (7) _G746 is 2+ -1 ? creep
^ Exit: (7) 1 is 2+ -1 ? creep
^ Call: (7) _G749 is 2+ -2 ? creep
^ Exit: (7) 0 is 2+ -2 ? creep
 Call: (7) fib(1, _G747) ? creep
 Exit: (7) fib(1, 1) ? creep
 Call: (7) fib(0, _G747) ? creep
 Exit: (7) fib(0, 0) ? creep
^ Call: (7) _G671 is 1+0 ? creep
^ Exit: (7) 1 is 1+0 ? creep
 Exit: (6) fib(2, 1) ? creep
M = 1 ;
 Redo: (7) fib(0, _G747) ? creep
^ Call: (8) _G752 is 0+ -1 ? creep
^ Exit: (8) -1 is 0+ -1 ? creep
^ Call: (8) _G755 is 0+ -2 ? creep
^ Exit: (8) -2 is 0+ -2 ? creep
 Call: (8) fib(-1, _G753) ? creep
^ Call: (9) _G758 is -1+ -1 ? creep
^ Exit: (9) -2 is -1+ -1 ? creep
^ Call: (9) _G761 is -1+ -2 ? creep
^ Exit: (9) -3 is -1+ -2 ? creep
 Call: (9) fib(-2, _G759) ? creep
^ Call: (10) _G764 is -2+ -1 ? creep
^ Exit: (10) -3 is -2+ -1 ? creep
...
</code></pre>

as you can see, after finding that the 2nd fibonacci is 1 (by your definition) you ask for a second solution; since you haven't specified that the third clause may only be used when N>1 prolog tries to find the second solution by calculating fib(-1), fib(-2), fib(-3) etc. 

to fix it, you have to add <code>N&gt;1</code> or a similar rule to the third clause

blocks|key|2067018|text|您可能要解决的一个问题是不必要地重新计算斐波那契数值。以下是对您的代码的一个小更改，以解决此缺陷：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2067019|:-+dynamic+db_fib/2.

init_fib+:-
++++assertz(+db_fib(0,+0)+),
++++assertz(+db_fib(1,+1)+).

fib(N,+NF)+:-
++++A+is+N+-+1,
++++B+is+N+-+2,
++++get_fib(A,+AF),
++++get_fib(B,+BF),
++++NF+is+AF+%2B+BF.

get_fib(A,+F)+:-
++++db_fib(A,+F),
++++!.

get_fib(A,+F)+:-
++++fib(A,+F),
++++assertz(+db_fib(A,+F)+).|code-block|syntax|javascript|2067020|例如，在SWI+Prolog中，可以计算|2067021|?-+init_fib,+fib(1000,F).|2067022|速度非常快，而且没有废气。|2067023|?-+init_fib.
true.

?-+fib(10,A).
A+=+55.

?-+fib(100,A).
A+=+354224848179261915075.

?-+fib(1000,A).
A+=+43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875.|2067024|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

One problem you might address is the unnecessary recomputation of fibonacci values. Here is a small change to your code to address this deficiency:

<pre><code>:- dynamic db_fib/2.

init_fib :-
 assertz( db_fib(0, 0) ),
 assertz( db_fib(1, 1) ).

fib(N, NF) :-
 A is N - 1,
 B is N - 2,
 get_fib(A, AF),
 get_fib(B, BF),
 NF is AF + BF.

get_fib(A, F) :-
 db_fib(A, F),
 !.

get_fib(A, F) :-
 fib(A, F),
 assertz( db_fib(A, F) ).
</code></pre>

For example, in SWI Prolog, it is possible to compute

<pre><code>?- init_fib, fib(1000,F).
</code></pre>

very quickly and with no stack exhaust.

<pre><code>?- init_fib.
true.

?- fib(10,A).
A = 55.

?- fib(100,A).
A = 354224848179261915075.

?- fib(1000,A).
A = 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875.
</code></pre>

blocks|key|2026593|text|您的代码不是tail-recursive.正确的尾递归结构意味着可以应用TRO+(尾递归优化)。这实际上是通过重用递归调用的现有堆栈帧，将递归转换为迭代。应用TRO后，每个递归调用都会在调用堆栈上推入一个新的堆栈框架。您应该像这样构造谓词(请注意，我还没有实际测试过这段代码，但它应该可以完成这项工作)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2026594|%25+------------------------------------------------------
%25+the+public+interface+predicate
%25+------------------------------------------------------
fib(1,1).++++++++++%25+first++element+in+the+sequence+is+1
fib(2,1).++++++++++%25+second+element+in+the+sequence+is+1
fib(N,X)+:-++++++++%25+subsequent+elements
++N+>+2+,++++++++++%25+++where+N+>+2
++fib(1,1,3,N,X)+++%25+++are+computed
++.

%25+--------------------------------------
%25+the+private+worker+predicate+for+N+>+2
%25+this+predicate+maintains+a+sliding+'window'+on+the+fibonacci+sequence
%25+as+it+computes+it
%25+--------------------------------------
fib(+V1+,+V2+,+N+,+N+,+X+)+:-+%25+compute+the+element+at+position+N
++N+>+2+,+++++++++++++++++++++%25+ensure+N+>+2
++X+is+V1+%2B+V2++++++++++++++++%25+value+is+the+sum+of+the+2+prior+elements
++.
fib(+V1+,+V2+,+T+,+N+,+X+)+:-+%25+on+backtracking,+slide+the+window+to+the+right:
++T+>+2+++++++++,+++++++++++++%25++ensure+N+>+2
++T1+is+T++%2B+1++,+++++++++++++%25++increment+N
++V3+is+V1+%2B+V2+,+++++++++++++%25++compute+the+next+value+(V1%2BV2)
++fib(V2,V3,T1,N,X)+++++++++++%25++recurse
++.|code-block|syntax|javascript|2026595|entityMap^0|6|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Your code isn't tail-recursive. a properly tail-recursive structure means that TRO (tail recursion optimization) can be applied. That essentially converts your recursion into iteration, by reusing the existing stack frame for the recursive call. With TRO applied, each recursive call pushes a new stack frame on the call stack. You should structure your predicate something like this (note that I haven't actually tested this code, but it should do the job):

<pre><code>% ------------------------------------------------------
% the public interface predicate
% ------------------------------------------------------
fib(1,1). % first element in the sequence is 1
fib(2,1). % second element in the sequence is 1
fib(N,X) :- % subsequent elements
 N &gt; 2 , % where N &gt; 2
 fib(1,1,3,N,X) % are computed
 .

% --------------------------------------
% the private worker predicate for N &gt; 2
% this predicate maintains a sliding 'window' on the fibonacci sequence
% as it computes it
% --------------------------------------
fib( V1 , V2 , N , N , X ) :- % compute the element at position N
 N &gt; 2 , % ensure N &gt; 2
 X is V1 + V2 % value is the sum of the 2 prior elements
 .
fib( V1 , V2 , T , N , X ) :- % on backtracking, slide the window to the right:
 T &gt; 2 , % ensure N &gt; 2
 T1 is T + 1 , % increment N
 V3 is V1 + V2 , % compute the next value (V1+V2)
 fib(V2,V3,T1,N,X) % recurse
 .
</code></pre>

blocks|key|2067054|text|很可能是这样的顺序(谁先是鸡还是蛋)，如果写成这样的话：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2067055|fib(N,+NF)+:-+
++A+is+N+-+1,+
++B+is+N+-+2,
++fib(A,+AF),+
++fib(B,+BF),
++NF+is+AF+%2B+BF.
fib(1,+1).
fib(0,+0).|code-block|syntax|javascript|2067056|这个问题将会得到解决。|2067057|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Is most likely the order(who is first the chicken or the egg) most likely if it was write like these:

<pre><code>fib(N, NF) :- 
 A is N - 1, 
 B is N - 2,
 fib(A, AF), 
 fib(B, BF),
 NF is AF + BF.
fib(1, 1).
fib(0, 0).
</code></pre>

the problem will be solve.

I wrote a predicate fib/2 to calculate the Fibonacci numbers in Prolog.
Though it works, it always says "out of local stack" and the error looks like:

<pre><code>?- fib(10, F).
F = 55 ;
ERROR: Out of local stack
</code></pre>

my predicate is below:

<pre><code>fib(0, 0).
fib(1, 1).
fib(N, NF) :-
 A is N - 1, 
 B is N - 2,
 fib(A, AF), 
 fib(B, BF),
 NF is AF + BF.
</code></pre>

Anyone knows why this is and how to fix it to get the following stuff::

<pre><code>% or the search might stop immediately, without pressing space.
?- fib2(10, F).
F = 55 ;
false. 
</code></pre>

Thanks in advance!

Why my predicate in Prolog Fib/2 always says "out of local stack"?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我用Prolog编写了一个谓词fib/2来计算斐波那契数。尽管它可以工作，但它总是显示"out of local stack“，错误看起来像这样：?- fib(10, F).F = 55 ;ERROR: Out of local stack我的谓词如下：fib(0, 0).fib(1, 1).fib(N, NF) :-...

问为什么我在Prolog Fib/2中的谓词总是说“在局部堆栈之外”？
EN

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问为什么我在Prolog Fib/2中的谓词总是说“在局部堆栈之外”？EN