如何驱动 C # 、 C + + 或 Java 编译器在编译时计算1 + 2 + 3 + ... + 1000？

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您可以使用(并且主要是滥用) C + + 宏/模板来执行元编程。 AFAIK，Java 不允许同样的东西。

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最佳答案

Updated Now with improved recursion depth! Works on MSVC10 and GCC without increased depth. :)

简单的编译时递归 + 附加:

template<unsigned Cur, unsigned Goal>
struct adder{
static unsigned const sub_goal = (Cur + Goal) / 2;
static unsigned const tmp = adder<Cur, sub_goal>::value;
static unsigned const value = tmp + adder<sub_goal+1, Goal>::value;
};


template<unsigned Goal>
struct adder<Goal, Goal>{
static unsigned const value = Goal;
};

Testcode:

template<unsigned Start>
struct sum_from{
template<unsigned Goal>
struct to{
template<unsigned N>
struct equals;


typedef equals<adder<Start, Goal>::value> result;
};
};


int main(){
sum_from<1>::to<1000>::result();
}

海湾合作委员会的产出:

错误: 声明‘ struct sum _ from < 1u > : : to < 1000u > : : equals < 500500u >

在 Ideone 上的现场示例。

MSVC10的输出:

error C2514: 'sum_from<Start>::to<Goal>::equals<Result>' : class has no constructors
with
[
Start=1,
Goal=1000,
Result=500500
]

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In theory, you can use this:

#include <iostream>


template<int N>
struct Triangle{
static int getVal()
{
return N + Triangle<N-1>::getVal();
}
};


template<>
struct Triangle<1>{
static int getVal()
{
return 1;
}
};


int main(){
std::cout << Triangle<1000>::getVal() << std::endl;
return 0;
}

但是海湾合作委员会给了我这个错误:

triangle.c++:7: error: template instantiation depth exceeds maximum of 500 (use -ftemplate-depth-NN to increase the maximum) instantiating struct Triangle<500>

再加上一个巨大的伪堆栈跟踪。

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我应该编写一个程序，可以驱动编译器在编译时计算这个和，并在编译完成时打印结果。

在编译期间打印数字的一个流行技巧是尝试访问一个模板的不存在成员，该模板实例化了您想要打印的数字:

template<int> struct print_n {};


print_n<1000 * 1001 / 2>::foobar go;

编译器接着说:

error: 'foobar' in 'struct print_n<500500>' does not name a type

有关此技术的更有趣的示例，请参见在编译时解决八个皇后的问题。

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C # 示例在编译时出错。

class Foo
{
const char Sum = (1000 + 1) * 1000 / 2;
}

产生以下编译错误:

Constant value '500500' cannot be converted to a 'char'

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下面是一个在 VC + + 2010下工作的实现。我不得不把计算分成3个阶段，因为当模板递归500多次时，编译器会抱怨。

template<int t_startVal, int t_baseVal = 0, int t_result = 0>
struct SumT
{
enum { result = SumT<t_startVal - 1, t_baseVal, t_baseVal + t_result +
t_startVal>::result };
};


template<int t_baseVal, int t_result>
struct SumT<0, t_baseVal, t_result>
{
enum { result = t_result };
};


template<int output_value>
struct Dump
{
enum { value = output_value };
int bad_array[0];
};


enum
{
value1 = SumT<400>::result,                // [1,400]
value2 = SumT<400, 400, value1>::result,   // [401, 800]
value3 = SumT<200, 800, value2>::result    // [801, 1000]
};


Dump<value3> dump;

当您编译此文件时，您应该看到编译器的输出如下所示:

1>warning C4200: nonstandard extension used : zero-sized array in struct/union
1>          Cannot generate copy-ctor or copy-assignment operator when UDT contains a
zero-sized array
1>          templatedrivensum.cpp(33) : see reference to class template
instantiation 'Dump<output_value>' being compiled
1>          with
1>          [
1>              output_value=500500
1>          ]

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由于面试问题中既没有指定编译器也没有指定语言，我敢在哈斯克尔提交一个使用 GHC 的解决方案:

{-# LANGUAGE TemplateHaskell #-}
{-# OPTIONS_GHC -ddump-splices #-}
module Main where


main :: IO ()
main = print $(let x = sum [1 :: Int .. 1000] in [| x |])

编译它:

$ ghc compsum.hs
[1 of 1] Compiling Main             ( compsum.hs, compsum.o )
Loading package ghc-prim ... linking ... done.
<snip more "Loading package ..." messages>
Loading package template-haskell ... linking ... done.
compsum.hs:6:16-56: Splicing expression
let x = sum [1 :: Int .. 1000] in [| x |] ======> 500500
Linking compsum ...

我们还有一个工作计划。

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C + + 11增加了 constexpr函数用于编译时计算，尽管目前只支持 gcc 4.6或更高版本，但使用 C + + 11会容易得多。

constexpr unsigned sum(unsigned start, unsigned end) {
return start == end ? start :
sum(start, (start + end) / 2) +
sum((start + end) / 2 + 1, end);
}


template <int> struct equals;
equals<sum(1,1000)> x;

该标准只要求编译器支持512的递归深度，所以它仍然需要避免线性递归深度。输出如下:

$ g++-mp-4.6 --std=c++0x test.cpp -c
test.cpp:8:25: error: aggregate 'equals<500500> x' has incomplete type and cannot be defined

当然你可以用这个公式:

constexpr unsigned sum(unsigned start, unsigned end) {
return (start + end) * (end - start + 1) / 2;
}


// static_assert is a C++11 assert, which checks
// at compile time.
static_assert(sum(0,1000) == 500500, "Sum failed for 0 to 1000");

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在 java 中，我考虑过使用注释处理。 Apt 工具在将源文件实际解析为 javac 命令之前扫描源文件。

在编制源文件期间，输出将打印出来:

@Documented
@Retention(RetentionPolicy.RUNTIME)
@Target({ElementType.TYPE, ElementType.METHOD})
public @interface MyInterface {


int offset() default 0;


int last() default 100;
}

The processor factory:

public class MyInterfaceAnnotationProcessorFactory implements AnnotationProcessorFactory {


public Collection<String> supportedOptions() {
System.err.println("Called supportedOptions.............................");
return Collections.EMPTY_LIST;
}


public Collection<String> supportedAnnotationTypes() {
System.err.println("Called supportedAnnotationTypes...........................");
return Collections.singletonList("practiceproject.MyInterface");
}


public AnnotationProcessor getProcessorFor(Set<AnnotationTypeDeclaration> set, AnnotationProcessorEnvironment ape) {
System.err.println("Called getProcessorFor................");
if (set.isEmpty()) {
return AnnotationProcessors.NO_OP;
}
return new MyInterfaceAnnotationProcessor(ape);
}
}

实际的注释处理器:

public class MyInterfaceAnnotationProcessor implements AnnotationProcessor {


private AnnotationProcessorEnvironment ape;
private AnnotationTypeDeclaration atd;


public MyInterfaceAnnotationProcessor(AnnotationProcessorEnvironment ape) {
this.ape = ape;
atd = (AnnotationTypeDeclaration) ape.getTypeDeclaration("practiceproject.MyInterface");
}


public void process() {
Collection<Declaration> decls = ape.getDeclarationsAnnotatedWith(atd);
for (Declaration dec : decls) {
processDeclaration(dec);
}
}


private void processDeclaration(Declaration d) {
Collection<AnnotationMirror> ams = d.getAnnotationMirrors();
for (AnnotationMirror am : ams) {
if (am.getAnnotationType().getDeclaration().equals(atd)) {
Map<AnnotationTypeElementDeclaration, AnnotationValue> values = am.getElementValues();
int offset = 0;
int last = 100;
for (Map.Entry<AnnotationTypeElementDeclaration, AnnotationValue> entry : values.entrySet()) {
AnnotationTypeElementDeclaration ated = entry.getKey();
AnnotationValue v = entry.getValue();
String name = ated.getSimpleName();
if (name.equals("offset")) {
offset = ((Integer) v.getValue()).intValue();
} else if (name.equals("last")) {
last = ((Integer) v.getValue()).intValue();
}
}
//find the sum
System.err.println("Sum: " + ((last + 1 - offset) / 2) * (2 * offset + (last - offset)));
}
}
}
}

然后我们创建一个源文件. simple 类，它使用 MyInterface 注释:

 @MyInterface(offset = 1, last = 1000)
public class Main {


@MyInterface
void doNothing() {
System.out.println("Doing nothing");
}


/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
Main m = new Main();
m.doNothing();
MyInterface my = (MyInterface) m.getClass().getAnnotation(MyInterface.class);
System.out.println("offset: " + my.offset());
System.out.println("Last: " + my.last());
}
}

注释处理器被编译成一个 jar 文件，然后使用 apt 工具将源文件编译为:

apt -cp "D:\Variance project\PracticeProject\dist\practiceproject.jar" -factory practiceproject.annotprocess.MyInterfaceAnnotationProcessorFactory "D:\Variance project\PracticeProject2\src\practiceproject2\Main.java"

The output of the project:

Called supportedAnnotationTypes...........................
Called getProcessorFor................
Sum: 5000
Sum: 500500

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I feel obligated to give this C code, since nobody else has yet:

#include <stdio.h>
int main() {
int x = 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+
21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+
41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+
61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+
81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100+
101+102+103+104+105+106+107+108+109+110+111+112+113+114+115+116+117+118+119+120+
121+122+123+124+125+126+127+128+129+130+131+132+133+134+135+136+137+138+139+140+
141+142+143+144+145+146+147+148+149+150+151+152+153+154+155+156+157+158+159+160+
161+162+163+164+165+166+167+168+169+170+171+172+173+174+175+176+177+178+179+180+
181+182+183+184+185+186+187+188+189+190+191+192+193+194+195+196+197+198+199+200+
201+202+203+204+205+206+207+208+209+210+211+212+213+214+215+216+217+218+219+220+
221+222+223+224+225+226+227+228+229+230+231+232+233+234+235+236+237+238+239+240+
241+242+243+244+245+246+247+248+249+250+251+252+253+254+255+256+257+258+259+260+
261+262+263+264+265+266+267+268+269+270+271+272+273+274+275+276+277+278+279+280+
281+282+283+284+285+286+287+288+289+290+291+292+293+294+295+296+297+298+299+300+
301+302+303+304+305+306+307+308+309+310+311+312+313+314+315+316+317+318+319+320+
321+322+323+324+325+326+327+328+329+330+331+332+333+334+335+336+337+338+339+340+
341+342+343+344+345+346+347+348+349+350+351+352+353+354+355+356+357+358+359+360+
361+362+363+364+365+366+367+368+369+370+371+372+373+374+375+376+377+378+379+380+
381+382+383+384+385+386+387+388+389+390+391+392+393+394+395+396+397+398+399+400+
401+402+403+404+405+406+407+408+409+410+411+412+413+414+415+416+417+418+419+420+
421+422+423+424+425+426+427+428+429+430+431+432+433+434+435+436+437+438+439+440+
441+442+443+444+445+446+447+448+449+450+451+452+453+454+455+456+457+458+459+460+
461+462+463+464+465+466+467+468+469+470+471+472+473+474+475+476+477+478+479+480+
481+482+483+484+485+486+487+488+489+490+491+492+493+494+495+496+497+498+499+500+
501+502+503+504+505+506+507+508+509+510+511+512+513+514+515+516+517+518+519+520+
521+522+523+524+525+526+527+528+529+530+531+532+533+534+535+536+537+538+539+540+
541+542+543+544+545+546+547+548+549+550+551+552+553+554+555+556+557+558+559+560+
561+562+563+564+565+566+567+568+569+570+571+572+573+574+575+576+577+578+579+580+
581+582+583+584+585+586+587+588+589+590+591+592+593+594+595+596+597+598+599+600+
601+602+603+604+605+606+607+608+609+610+611+612+613+614+615+616+617+618+619+620+
621+622+623+624+625+626+627+628+629+630+631+632+633+634+635+636+637+638+639+640+
641+642+643+644+645+646+647+648+649+650+651+652+653+654+655+656+657+658+659+660+
661+662+663+664+665+666+667+668+669+670+671+672+673+674+675+676+677+678+679+680+
681+682+683+684+685+686+687+688+689+690+691+692+693+694+695+696+697+698+699+700+
701+702+703+704+705+706+707+708+709+710+711+712+713+714+715+716+717+718+719+720+
721+722+723+724+725+726+727+728+729+730+731+732+733+734+735+736+737+738+739+740+
741+742+743+744+745+746+747+748+749+750+751+752+753+754+755+756+757+758+759+760+
761+762+763+764+765+766+767+768+769+770+771+772+773+774+775+776+777+778+779+780+
781+782+783+784+785+786+787+788+789+790+791+792+793+794+795+796+797+798+799+800+
801+802+803+804+805+806+807+808+809+810+811+812+813+814+815+816+817+818+819+820+
821+822+823+824+825+826+827+828+829+830+831+832+833+834+835+836+837+838+839+840+
841+842+843+844+845+846+847+848+849+850+851+852+853+854+855+856+857+858+859+860+
861+862+863+864+865+866+867+868+869+870+871+872+873+874+875+876+877+878+879+880+
881+882+883+884+885+886+887+888+889+890+891+892+893+894+895+896+897+898+899+900+
901+902+903+904+905+906+907+908+909+910+911+912+913+914+915+916+917+918+919+920+
921+922+923+924+925+926+927+928+929+930+931+932+933+934+935+936+937+938+939+940+
941+942+943+944+945+946+947+948+949+950+951+952+953+954+955+956+957+958+959+960+
961+962+963+964+965+966+967+968+969+970+971+972+973+974+975+976+977+978+979+980+
981+982+983+984+985+986+987+988+989+990+991+992+993+994+995+996+997+998+999+1000;
printf("%d\n", x);
}

我所要做的就是检查组装，找到我的答案！

gcc -S compile_sum.c;
grep "\$[0-9]*, *-4" compile_sum.s

And I see:

movl    $500500, -4(%rbp)

小开

使用 java，你可以对 C # 回答做类似的事情:

public class Cheat {
public static final int x = (1000 *1001/2);
}


javac -Xprint Cheat.java


public class Cheat {


public Cheat();
public static final int x = 500500;
}

you can do this in 使用 Peano 数字的 scala because you can force the compiler to do recursion but i don't think you can do the same thing in c#/java

另一个不使用 Xprint 的解决方案，但是更加狡猾

public class Cheat {
public static final int x = 5/(1000 *1001/2 - 500500);
}


javac -Xlint:all Cheat.java


Cheat.java:2: warning: [divzero] division by zero
public static final int x = 5/(1000 *1001/2 - 500500);
^
1 warning

不使用任何编译器标志。因为您可以检查任意数量的常量(而不仅仅是500500) ，所以这个解决方案应该是可以接受的。

public class Cheat {
public static final short max = (Short.MAX_VALUE - 500500) + 1001*1000/2;
public static final short overflow = (Short.MAX_VALUE - 500500 + 1) + 1001*1000/2;


}


Cheat.java:3: error: possible loss of precision
public static final short overflow = (Short.MAX_VALUE - 500500 + 1) + 1001*1000/2;
^
required: short
found:    int
1 error

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从卡尔 · 沃尔什的答案延伸到实际打印编译过程中的结果:

#define VALUE (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+\
21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+\
41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60+\
61+62+63+64+65+66+67+68+69+70+71+72+73+74+75+76+77+78+79+80+\
81+82+83+84+85+86+87+88+89+90+91+92+93+94+95+96+97+98+99+100+\
101+102+103+104+105+106+107+108+109+110+111+112+113+114+115+116+117+118+119+120+\
121+122+123+124+125+126+127+128+129+130+131+132+133+134+135+136+137+138+139+140+\
141+142+143+144+145+146+147+148+149+150+151+152+153+154+155+156+157+158+159+160+\
161+162+163+164+165+166+167+168+169+170+171+172+173+174+175+176+177+178+179+180+\
181+182+183+184+185+186+187+188+189+190+191+192+193+194+195+196+197+198+199+200+\
201+202+203+204+205+206+207+208+209+210+211+212+213+214+215+216+217+218+219+220+\
221+222+223+224+225+226+227+228+229+230+231+232+233+234+235+236+237+238+239+240+\
241+242+243+244+245+246+247+248+249+250+251+252+253+254+255+256+257+258+259+260+\
261+262+263+264+265+266+267+268+269+270+271+272+273+274+275+276+277+278+279+280+\
281+282+283+284+285+286+287+288+289+290+291+292+293+294+295+296+297+298+299+300+\
301+302+303+304+305+306+307+308+309+310+311+312+313+314+315+316+317+318+319+320+\
321+322+323+324+325+326+327+328+329+330+331+332+333+334+335+336+337+338+339+340+\
341+342+343+344+345+346+347+348+349+350+351+352+353+354+355+356+357+358+359+360+\
361+362+363+364+365+366+367+368+369+370+371+372+373+374+375+376+377+378+379+380+\
381+382+383+384+385+386+387+388+389+390+391+392+393+394+395+396+397+398+399+400+\
401+402+403+404+405+406+407+408+409+410+411+412+413+414+415+416+417+418+419+420+\
421+422+423+424+425+426+427+428+429+430+431+432+433+434+435+436+437+438+439+440+\
441+442+443+444+445+446+447+448+449+450+451+452+453+454+455+456+457+458+459+460+\
461+462+463+464+465+466+467+468+469+470+471+472+473+474+475+476+477+478+479+480+\
481+482+483+484+485+486+487+488+489+490+491+492+493+494+495+496+497+498+499+500+\
501+502+503+504+505+506+507+508+509+510+511+512+513+514+515+516+517+518+519+520+\
521+522+523+524+525+526+527+528+529+530+531+532+533+534+535+536+537+538+539+540+\
541+542+543+544+545+546+547+548+549+550+551+552+553+554+555+556+557+558+559+560+\
561+562+563+564+565+566+567+568+569+570+571+572+573+574+575+576+577+578+579+580+\
581+582+583+584+585+586+587+588+589+590+591+592+593+594+595+596+597+598+599+600+\
601+602+603+604+605+606+607+608+609+610+611+612+613+614+615+616+617+618+619+620+\
621+622+623+624+625+626+627+628+629+630+631+632+633+634+635+636+637+638+639+640+\
641+642+643+644+645+646+647+648+649+650+651+652+653+654+655+656+657+658+659+660+\
661+662+663+664+665+666+667+668+669+670+671+672+673+674+675+676+677+678+679+680+\
681+682+683+684+685+686+687+688+689+690+691+692+693+694+695+696+697+698+699+700+\
701+702+703+704+705+706+707+708+709+710+711+712+713+714+715+716+717+718+719+720+\
721+722+723+724+725+726+727+728+729+730+731+732+733+734+735+736+737+738+739+740+\
741+742+743+744+745+746+747+748+749+750+751+752+753+754+755+756+757+758+759+760+\
761+762+763+764+765+766+767+768+769+770+771+772+773+774+775+776+777+778+779+780+\
781+782+783+784+785+786+787+788+789+790+791+792+793+794+795+796+797+798+799+800+\
801+802+803+804+805+806+807+808+809+810+811+812+813+814+815+816+817+818+819+820+\
821+822+823+824+825+826+827+828+829+830+831+832+833+834+835+836+837+838+839+840+\
841+842+843+844+845+846+847+848+849+850+851+852+853+854+855+856+857+858+859+860+\
861+862+863+864+865+866+867+868+869+870+871+872+873+874+875+876+877+878+879+880+\
881+882+883+884+885+886+887+888+889+890+891+892+893+894+895+896+897+898+899+900+\
901+902+903+904+905+906+907+908+909+910+911+912+913+914+915+916+917+918+919+920+\
921+922+923+924+925+926+927+928+929+930+931+932+933+934+935+936+937+938+939+940+\
941+942+943+944+945+946+947+948+949+950+951+952+953+954+955+956+957+958+959+960+\
961+962+963+964+965+966+967+968+969+970+971+972+973+974+975+976+977+978+979+980+\
981+982+983+984+985+986+987+988+989+990+991+992+993+994+995+996+997+998+999+1000)


char tab[VALUE];


int main()
{
tab = 5;
}

Gcc 产出:

test.c: In function 'main':
test.c:56:9: error: incompatible types when assigning to type 'char[500500]' fro
m type 'int'

小开

虽然这实际上适用于小数，但是如果我使用 sum _ first，其中 N > 400，clang + + 返回一个编译器错误。

#include <iostream>


using namespace std;




template <int N>
struct sum_first
{
static const int value = N + sum_first<N - 1>::value;
};


template <>
struct sum_first<0>
{
static const int value = 0;
};


int main()
{
cout << sum_first<1000>::value << endl;
}