C ++ quine [英] C++ quine

问题描述

我很无聊，所以我只是在100％ansi c ++中编写了一个quine（程序将打印出它自己的

源代码），而不使用任何c-library

函数（例如printf），并且不对字符进行任何假设

设置。

有何评论？错误？关于如何缩短它的想法？


Emlyn

#include< string>

#include< iostream>

使用命名空间std;

int main（）{

string s =" s"，s0 =" \\\
"，s1 = " \""，s2 =" \\"，s3 =" n"，s4 =" ;;" ;;

string s5 =" string s"，s6 ="，s"，s7 =" ="，s8 = s7 + s1，s9 =" +";

string s10 =" #include< string>"，s11 =" #include< iostream> " ;;

string s12 =" using namespace std;"，s13 =" int main（）{" ;;

string s14 =" return 0;"，s15 ="}" ;;

string s20 =" COUT<< S10<< S0<< S11<< S0<< S12<< S0<< S13<< S0;英寸;

串S21 = QUOT;

COUT<< S5<< S8<< S<< S1和S2;< S6<&℃，LT;< S8<< S2<< S3<< S1和S2; < S6<&。1 LT; < s8<< s2<< s1<< s1<<< s6;" ;;

string s22 ="

cout<<< 2< < S8<< S2<< S2<< S1和S2;< S6<&3; LT;< S8<< S3<< S1和S2;< S6<< 4℃; < s8<< s4<< s1<< s4<< s0"" ;;

string s23 ="

cout<< s5<小于5<< S8<< S5<< S1和S2;< S6<&6; LT;< S8<< S6<< S1和S2;< S6<&7; LT; < s8<<< s7<< s1;" ;;

string s24 ="

cout<< s6<< 8<< s7< < S<&7; LT;< S9<< S<&。1 LT;< S6<&其中R 9 LT;< S8<< S9<< S1和S2;< S4<< S0;" ;;

string s25 =" COUT<< S5<< 10<< S8<< S10<< S1和S2;< S6<< 11<< S8<< S11<< S1和S2;< S4<< S0 ;" ;;

string s26 =" COUT<< S5<&其中R 12 LT;< S8<< S12<< S1和S2;< S6<&; 13 LT;< S8<< S13<< S1和S2;< S4<< S0 ;" ;;

string s27 =" COUT<< S5<< 14<< S8<< S14<< S1和S2;< S6<< 15℃;< S8<< S15<< S1和S2;< S4<< S0 ;英寸;

串S28 = QUOT;

COUT<< S5<< 20℃;< S8<< S20<< S1和S2;< S4< ;< S0<< S5<< 21<< S8<< S21<< S1和S2;< S4<< S0<< S5<< 22℃;< S8<< <无线电通信/>
s22<< s1<< s4<< s0;" ;;

string s29 ="

cout<< s5< < 23℃;< S8<< S23<< S1和S2;< S4<< S0<< S5<< 24<< S8<< S24<< S1和S2;< S4<< ; S0<< S5<< 25℃;< S8<<

S25<< S1和S2;< S4<< S0;英寸;

串S30 = QUOT;

COUT<< S5<< 26<< S8<< S26<< S1和S2;< S4<< S0<< S5<< 27< ;< S8<< S27<< S1和S2;< S4<< S0<< S5<< 28和60;< S8<<

S28<< S1和S2; < s4<<< s0;" ;;

string S31 = QUOT;

COUT<< S5<< 29<< S8<< S29<< S1和S2;< S4<< S0<< S5<< 30℃; < S8<< S30<< S1和S2;< S4<< S0<< S5<< 31<< S8<<

S31<< S1和S2;< ; s4<<<< s0;" ;;

string s32 ="

cout<< s5<< 32<< s8<< s32<< S1和S2;< S4<< S0<< S5<&γ-33 LT;< S8<< S33<< S1和S2;< S4<< S0<< S5<< 34 - ;< s8<<

s34<< s1<< s4<< s0;" ;;

string s33 ="

COUT<< S5<< 35℃;< S8<< S35<< S1和S2;< S4<< S0<< S5<&R 36取代LT;< S8<< S36<< S1和S2 ;< s4<<<< s0;" ;;

string s34 ="

cout<< s20<< s0<< s21<<< s0< ;< S22<< S0<< S23<< S0<< S24<< S0< < S25<< S0<< S26<< S0<< S27<< S0

将< S28<< S0;英寸;

串S35 = QUOT;

COUT<< S29<< S0<< S30<< S0<< S31<< S0<< S32<< S0<< ; S33<< S0< < s34<<< s0<< s35<< s0<< s36<< s0

;" ;;

string s36 =" COUT<< S14<< S0<< S15<< S0;英寸;

COUT<< S10<< S0<< S11<< S0<< S12< ;< S0<< S13<< S0;

COUT<< S5<< S8<< S<< S1和S2;< S6<&℃，LT;< S8< ;< S2<< S3<< S1和S2;< S6<&。1 LT; < S8<< S2<< S1和S2;< S1和S2;< S6;

COUT<< 2';< S8<< S2<< S2<< S1和S2; < S6<&3; LT;< S8<< S3<< S1和S2;< S6<< 4℃; < S8<< S4<< S1和S2;< S4<< S0;

COUT<< S5<< 5';< S8<< S5<< S1和S2; < S6<&6; LT;< S8<< S6<< S1和S2;< S6<&7; LT; < S8<< S7<< S1;

COUT<< S6<< 8 LT;< S7<< S<&7; LT;< S9<< S< &。1 LT;< S6<&其中R 9 LT;< S8<< S9<< S1和S2;< S4<< S0;

COUT<< S5<< 10< ;< S8<< S10<< S1和S2;< S6<< 11<< S8<< S11<< S1和S2;< S4<< S0;

COUT<< S5<&其中R 12 LT;< S8<< S12<< S1和S2;< S6<&; 13 LT;< S8<< S13<< S1和S2;< S4<< S0 ;

COUT<< S5<< 14<< S8<< S14<< S1和S2;< S6<< 15℃;< S8<< S15<< S1和S2 ;< S4<< S0;


COUT<< S5<< 20℃;< S8<< S20<< S1和S2;< S4<< S0<< S5<< 21<< S8<< S21<< S1和S2;< S4<< S0<< S5<< 22℃;< S8<<

S22<< S1和S2;< S4<< S0;


COUT<< S5<< 23℃;< S8<< S23<< S1和S2 ;< S4<< S0<< S5<< 24<< S8<< S24 << S1和S2;< S4<< S0<< S5<< 25℃;< S8<<

S25<< S1和S2;< S4<< S0;


COUT<< S5<< 26<< S8<< S26<< S1和S2;< S4<< S0<< S5<< 27< < S8<< S27<< S1和S2;< S4<< S0<< S5<< 28和60;< S8<<

S28<< S1和S2;< ; S4<< S0;


COUT<< S5<< 29<< S8<< S29<< S1和S2;< S4<< S0<< ; S5<< 30℃;< S8<< S30<< S1和S2;< S4<< S0<< S5<< 31<< S8<<

S31<< S1和S2;< S4<< S0;


COUT<< S5<&γ-32 LT;< S8<< S32<< S1和S2;< S4<< S0<< S5<&γ-33 LT;< S8<< S33<< S1和S2;< S4<< S0<< S5<< 34 - ;< S8<< <登记/>
S34<< S1和S2;< S4<< S0;

COUT<< S5<< 35℃;< S8<< S35<< S1和S2 ;< s4<<< s0<< s5<<<<<< s<<<< s<吨;< S1和S2;< S4<< S0;


COUT<< S20<< S0<< S21<< S0<< S22<< S0<< S23<< S0<< S24<< S0< < s25<<< s0<< s26<< s0<< s27<< s0

<< s28<< s0;


COUT<< S29<< S0<< S30<< S0<< S31<< S0<< S32<< S0<< S33<< S0< < S34<< S0<< S35<< S0<< S36<< S0

;

COUT<< S14<< S0< < s15<< s0;

返回0;

}

I was bored, so I just wrote a quine (program that will print out it''s own
source code when run) in 100% ansi c++, without using any c-library
functions (such as printf), and not making any assumptions on the character
set.
Any comments? bugs? ideas on how to shorten it?

Emlyn
#include <string>
#include <iostream>
using namespace std;
int main() {
string s = "s", s0 = "\n", s1 = "\"", s2 = "\\", s3 = "n", s4 = ";";
string s5 = " string s", s6 = ", s", s7 = " = ", s8 = s7+s1, s9 = "+";
string s10 = "#include <string>", s11 = "#include <iostream>";
string s12 = "using namespace std;", s13 = "int main() {";
string s14 = " return 0;", s15 = "}";
string s20 = " cout<<s10<<s0<<s11<<s0<<s12<<s0<<s13<<s0;";
string s21 = "
cout<<s5<<s8<<s<<s1<<s6<<0<<s8<<s2<<s3<<s1<<s6<<1< <s8<<s2<<s1<<s1<<s6;";
string s22 = "
cout<<2<<s8<<s2<<s2<<s1<<s6<<3<<s8<<s3<<s1<<s6<<4< <s8<<s4<<s1<<s4<<s0;";
string s23 = "
cout<<s5<<5<<s8<<s5<<s1<<s6<<6<<s8<<s6<<s1<<s6<<7< <s8<<s7<<s1;";
string s24 = "
cout<<s6<<8<<s7<<s<<7<<s9<<s<<1<<s6<<9<<s8<<s9<<s1 <<s4<<s0;";
string s25 = " cout<<s5<<10<<s8<<s10<<s1<<s6<<11<<s8<<s11<<s1<<s4 <<s0;";
string s26 = " cout<<s5<<12<<s8<<s12<<s1<<s6<<13<<s8<<s13<<s1<<s4 <<s0;";
string s27 = " cout<<s5<<14<<s8<<s14<<s1<<s6<<15<<s8<<s15<<s1<<s4 <<s0;";
string s28 = "
cout<<s5<<20<<s8<<s20<<s1<<s4<<s0<<s5<<21<<s8<<s21 <<s1<<s4<<s0<<s5<<22<<s8<<
s22<<s1<<s4<<s0;";
string s29 = "
cout<<s5<<23<<s8<<s23<<s1<<s4<<s0<<s5<<24<<s8<<s24 <<s1<<s4<<s0<<s5<<25<<s8<<
s25<<s1<<s4<<s0;";
string s30 = "
cout<<s5<<26<<s8<<s26<<s1<<s4<<s0<<s5<<27<<s8<<s27 <<s1<<s4<<s0<<s5<<28<<s8<<
s28<<s1<<s4<<s0;";
string s31 = "
cout<<s5<<29<<s8<<s29<<s1<<s4<<s0<<s5<<30<<s8<<s30 <<s1<<s4<<s0<<s5<<31<<s8<<
s31<<s1<<s4<<s0;";
string s32 = "
cout<<s5<<32<<s8<<s32<<s1<<s4<<s0<<s5<<33<<s8<<s33 <<s1<<s4<<s0<<s5<<34<<s8<<
s34<<s1<<s4<<s0;";
string s33 = "
cout<<s5<<35<<s8<<s35<<s1<<s4<<s0<<s5<<36<<s8<<s36 <<s1<<s4<<s0;";
string s34 = "
cout<<s20<<s0<<s21<<s0<<s22<<s0<<s23<<s0<<s24<<s0< <s25<<s0<<s26<<s0<<s27<<s0
<<s28<<s0;";
string s35 = "
cout<<s29<<s0<<s30<<s0<<s31<<s0<<s32<<s0<<s33<<s0< <s34<<s0<<s35<<s0<<s36<<s0
;";
string s36 = " cout<<s14<<s0<<s15<<s0;";
cout<<s10<<s0<<s11<<s0<<s12<<s0<<s13<<s0;
cout<<s5<<s8<<s<<s1<<s6<<0<<s8<<s2<<s3<<s1<<s6<<1< <s8<<s2<<s1<<s1<<s6;
cout<<2<<s8<<s2<<s2<<s1<<s6<<3<<s8<<s3<<s1<<s6<<4< <s8<<s4<<s1<<s4<<s0;
cout<<s5<<5<<s8<<s5<<s1<<s6<<6<<s8<<s6<<s1<<s6<<7< <s8<<s7<<s1;
cout<<s6<<8<<s7<<s<<7<<s9<<s<<1<<s6<<9<<s8<<s9<<s1 <<s4<<s0;
cout<<s5<<10<<s8<<s10<<s1<<s6<<11<<s8<<s11<<s1<<s4 <<s0;
cout<<s5<<12<<s8<<s12<<s1<<s6<<13<<s8<<s13<<s1<<s4 <<s0;
cout<<s5<<14<<s8<<s14<<s1<<s6<<15<<s8<<s15<<s1<<s4 <<s0;

cout<<s5<<20<<s8<<s20<<s1<<s4<<s0<<s5<<21<<s8<<s21 <<s1<<s4<<s0<<s5<<22<<s8<<
s22<<s1<<s4<<s0;

cout<<s5<<23<<s8<<s23<<s1<<s4<<s0<<s5<<24<<s8<<s24 <<s1<<s4<<s0<<s5<<25<<s8<<
s25<<s1<<s4<<s0;

cout<<s5<<26<<s8<<s26<<s1<<s4<<s0<<s5<<27<<s8<<s27 <<s1<<s4<<s0<<s5<<28<<s8<<
s28<<s1<<s4<<s0;

cout<<s5<<29<<s8<<s29<<s1<<s4<<s0<<s5<<30<<s8<<s30 <<s1<<s4<<s0<<s5<<31<<s8<<
s31<<s1<<s4<<s0;

cout<<s5<<32<<s8<<s32<<s1<<s4<<s0<<s5<<33<<s8<<s33 <<s1<<s4<<s0<<s5<<34<<s8<<
s34<<s1<<s4<<s0;
cout<<s5<<35<<s8<<s35<<s1<<s4<<s0<<s5<<36<<s8<<s36 <<s1<<s4<<s0;

cout<<s20<<s0<<s21<<s0<<s22<<s0<<s23<<s0<<s24<<s0< <s25<<s0<<s26<<s0<<s27<<s0
<<s28<<s0;

cout<<s29<<s0<<s30<<s0<<s31<<s0<<s32<<s0<<s33<<s0< <s34<<s0<<s35<<s0<<s36<<s0
;
cout<<s14<<s0<<s15<<s0;
return 0;
}

推荐答案

Emlyn> ;我很无聊，所以我只是写了一个quine（程序将在100％ansi c ++中打印出来，然后运行它自己的源代码），没有

Emlyn>使用任何c库函数（如printf），而不是制作

Emlyn>关于字符集的任何假设。


Emlyn>任何意见？错误？关于如何缩短它的想法？


这是一个C89版本，用我觉得更容易理解的形式写成。感兴趣的读者可能喜欢将其翻译成C ++：


#include< stdio.h>


char * self [] = {

" #include< stdio.h>"，

""，

" char * self [] = { "，

" \ n"，

" \t0"，

"};"，

""，

" main（）"，

" {"，

" ; \ tchar * p，** s;"，

""，

" \ tfor（s = self; * s; s ++） {"，

" \t\tif（** s ==''\\ n''）{"，

" \ tt \ tttttt;"，

" \t \ t \ tt（t = self; * t; t ++）{"，

" \t\t\t\tputchar（''\\\\''）;"，

" \ t \\ \\t \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ * p; p ++）{&q uot;，

" \t\t\t\t\tif（* p ==''\\ t''）{"，

" \t\t\t\t\t\tputchar（''\\\\''）;"，

\ t \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\如果（* p ==''\"''）{"，

" \t \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \t\tputchar（''\\\\''）;"，

" \t\t\t\t\\\ \\ tttcharchar（''\"''）;"，

" \t\t\t\t\t}否则if（* p ==' '\\ n''）{"，

" \t\t\t\t\t\tputchar（''\\\ \''）;"，

" \t\t\t\t\t\tputchar（''n''）;"，

" \t\t\t\t\t}否则如果（* p ==''\\\\''）{"，

" \t\t\t\t\t\tputchar （''\\\\''）;"，

" \t\t\t\t\t\tputchar（''\\' \\\\\''）;"，

" \t\t\t\t\t} else"，

" \t\t\t\t\t\tputchar（* p）;"，

" \t\t\\ t}"，

" \t\t\t\tputchar（''\"''）;"，

" ; \t\t\t\tputchar（''，''）;"，

" \t\t\t\tputchar（''\\ \\\ n''）;"，

" \t\t\t}"，

" \t\ t } else {"，

" \t\t\tfor（p = * s; * P; p ++）{"，

" \t\t\t\tputchar（* p）;"，

" \ t \\ \\ t \}，

" \t\t\tputchar（''\\ n''）;"，

" \t\t}"，

" \t}"，

" \treturn 0;"，

"}"，

0

};


main（）

{

char * p，** s;


for（s = self; * s; s ++）{

if（** s ==''\ n''）{

char ** t;

for（t = self; * t; t ++ ）{

putchar（''\t''）;

putchar（''"''）;

for（p = * t; * p; p ++）{

if（* p ==''\t''）{

putchar（''\\' '）;

putchar（''t''）;

}否则if（* p ==''"''）{

putchar（''\\''）;

putchar（''"''）;

}否则if（* p ==' '\ n''）{

putchar（''\\''）;

putchar（''n''）;

}否则if（* p ==''\\''）{

putchar（''\\''）;

putchar （''\\''）;

}其他

putchar（* p）;

}

putchar（''"''）;

putchar（''，''）;

putchar（''\ n''）;

}

}否则{

for（p = * s; * P; p ++）{

putchar（* p）;

}

putchar（''\ n''）;

}

}

返回0;

}

-

Andrew Koenig， ar*@acm.org

Emlyn> I was bored, so I just wrote a quine (program that will print
Emlyn> out it''s own source code when run) in 100% ansi c++, without
Emlyn> using any c-library functions (such as printf), and not making
Emlyn> any assumptions on the character set.

Emlyn> Any comments? bugs? ideas on how to shorten it?

Here''s a C89 version, written in a form that I find easier to
understand. The interested reader may enjoy translating it to C++:

#include <stdio.h>

char *self[] = {
"#include <stdio.h>",
"",
"char *self[] = {",
"\n",
"\t0",
"};",
"",
"main()",
"{",
"\tchar *p, **s;",
"",
"\tfor (s = self; *s; s++) {",
"\t\tif (**s == ''\\n'') {",
"\t\t\tchar **t;",
"\t\t\tfor (t = self; *t; t++) {",
"\t\t\t\tputchar (''\\t'');",
"\t\t\t\tputchar (''\"'');",
"\t\t\t\tfor (p = *t; *p; p++) {",
"\t\t\t\t\tif (*p == ''\\t'') {",
"\t\t\t\t\t\tputchar (''\\\\'');",
"\t\t\t\t\t\tputchar (''t'');",
"\t\t\t\t\t} else if (*p == ''\"'') {",
"\t\t\t\t\t\tputchar (''\\\\'');",
"\t\t\t\t\t\tputchar (''\"'');",
"\t\t\t\t\t} else if (*p == ''\\n'') {",
"\t\t\t\t\t\tputchar (''\\\\'');",
"\t\t\t\t\t\tputchar (''n'');",
"\t\t\t\t\t} else if (*p == ''\\\\'') {",
"\t\t\t\t\t\tputchar (''\\\\'');",
"\t\t\t\t\t\tputchar (''\\\\'');",
"\t\t\t\t\t} else",
"\t\t\t\t\t\tputchar (*p);",
"\t\t\t\t}",
"\t\t\t\tputchar (''\"'');",
"\t\t\t\tputchar ('','');",
"\t\t\t\tputchar (''\\n'');",
"\t\t\t}",
"\t\t} else {",
"\t\t\tfor (p = *s; *p; p++) {",
"\t\t\t\tputchar (*p);",
"\t\t\t}",
"\t\t\tputchar (''\\n'');",
"\t\t}",
"\t}",
"\treturn 0;",
"}",
0
};

main()
{
char *p, **s;

for (s = self; *s; s++) {
if (**s == ''\n'') {
char **t;
for (t = self; *t; t++) {
putchar (''\t'');
putchar (''"'');
for (p = *t; *p; p++) {
if (*p == ''\t'') {
putchar (''\\'');
putchar (''t'');
} else if (*p == ''"'') {
putchar (''\\'');
putchar (''"'');
} else if (*p == ''\n'') {
putchar (''\\'');
putchar (''n'');
} else if (*p == ''\\'') {
putchar (''\\'');
putchar (''\\'');
} else
putchar (*p);
}
putchar (''"'');
putchar ('','');
putchar (''\n'');
}
} else {
for (p = *s; *p; p++) {
putchar (*p);
}
putchar (''\n'');
}
}
return 0;
}
--
Andrew Koenig, ar*@acm.org

Emlyn Corrin写道：

Emlyn Corrin wrote:

我很无聊，所以我只是在100％ansi c ++中编写了一个quine（程序将在运行时打印出它自己的源代码），而不使用任何c-library
函数（如printf），而不对字符集进行任何假设。
有何评论？错误？关于如何缩短它的想法？

Emlyn

I was bored, so I just wrote a quine (program that will print out it''s own
source code when run) in 100% ansi c++, without using any c-library
functions (such as printf), and not making any assumptions on the character
set.
Any comments? bugs? ideas on how to shorten it?

Emlyn

这是我的尝试:)我很确定这是标准的C ++，我还有

怀疑那些拥有更多C ++知识的人可以缩短它的时间。

me。如果这被你的新闻阅读器吃掉，请注意我没有

多行字面意思:)


唯一的小问题是我怀疑我做不了

b.insert（它，''\\''），因为这会使它无效（这对我来说意味着

之后插入物品会很明智*叹气*）

，所以我应该制作一个临时的减量副本，但我不能

打扰:)我也可以硬连接a.insert线和b.reserve

线，但是现在想让它们保持通用:)


#include< string>

#include< iostream>

using namespace std;

string a（" #include< string> \ n＃include< iostream> \ nusing namespace

std \ nstring a（\" X \"）int main（）{\ nstring

b = a; \ nb.reserve（b.length（）* 2）; \ nstring :: iterator

it = b.begin（）; \ nwhile（it！= b.end（））{\ nif（* it ==''\ n''）{* it =''n'';

b.insert（it，''\\ \\\''）;} \ nit ++; \ n} \ nna.insert（find（ab egin（），a.end（），''X''），b.begin（ ），b.end（））; \\\
cout<< A<< endl; \ n} \ n"）;

int main（）{

string b = a;

b.reserve（ b.length（）* 2）;

string :: iterator it = b.begin（）;

while（it！= b.end（））{

if（* it ==''\ n''）{* it =''n''; b.insert（it，''\\''）;}

it ++;

}

a.insert（find（ a.begin（），a.end（），''X''），b.begin（），be nd（））;

cout<< a<< endl;

}

Here is my attempt :) I''m fairly certain is this standard C++, I also
suspect it can be made shorter by someone with more C++ knowledge than
me. If this gets eaten by your newsreader, note that I don''t have a
multi-line literal :)

The only minor point is that I suspect that I can''t do
b.insert(it,''\\''), as this will invalidate it (which suggests to me it
would have been sensible to have insert put things in afterwards *sigh*)
, so I should perhaps make a temporary decremented copy, but I can''t be
bothered :) Also I could hard-wire the a.insert line, and the b.reserve
line, but wanted to keep them general for now :)

#include<string>
#include<iostream>
using namespace std;
string a("#include<string>\n#include<iostream>\nusing namespace
std\nstring a(\"X\")int main(){\nstring
b=a;\nb.reserve(b.length()*2);\nstring::iterator
it=b.begin();\nwhile(it!=b.end()) {\nif(*it==''\n'') {*it=''n'';
b.insert(it,''\\\\'');}\nit++;\n}\na.insert(find(a.b egin(),a.end(),''X''),b.begin(),b.end());\ncout<<a<< endl;\n}\n");
int main(){
string b=a;
b.reserve(b.length()*2);
string::iterator it=b.begin();
while(it!=b.end()) {
if(*it==''\n'') {*it=''n''; b.insert(it,''\\'');}
it++;
}
a.insert(find(a.begin(),a.end(),''X''),b.begin(),b.e nd());
cout<<a<<endl;
}

我希望新闻阅读器不要太过混乱。

f的定义没有换行符。


Owner @ machine~ / projects / quine

I hope the newsreader doesn''t mess this up too much. There are no newlines in
the definition of f.

Owner@machine ~/projects/quine

这篇关于C ++ quine的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！