有什么方法可以使这段代码更紧凑，和/或能够在运行时更改？ [英] Any way to make this code more compact, and/or be able to change at runtime?
问题描述
我写的这个程序模拟了一个36个字符，10个转子
互易转子密码，带有一个插板。我有什么方法可以使
插板功能更紧凑和/或能够在运行时更改映射
？
char Enigma :: plugboard（char char）
{
if（Char ==''A''）
返回''0' ';
否则if（Char ==''B''）
返回''Q'';
else if（Char = ='''C''）
返回''W'';
否则if（Char ==''D''）
返回''E'';
否则if（Char ==''E''）
返回''D'';
否则if（Char ==''F''）
返回''T'';
else if（Char ==''G''）
返回''Y'';
否则if（Char ==''H''）
返回''U'';
else if（Char ==''I''）
返回''我';
else if（Char ==''J' '）
返回''O'';
否则if（Char ==''K''）
返回''P' ';
否则if（Char ==''L''）
返回''S'';
else if（Char ==''M''）
返回''N'';
否则if（Char ==''N''）
返回''M'';
else if（Char ==''O''）
返回''J'';
else if（Char ==''P' '）
返回''K'';
否则if（Char ==''Q''）
返回''B' ';
否则if（Char ==''R''）
返回''Z'';
else if（Char = =''S''）
返回''L'';
否则if（Char ==''T''）
返回''F'';
否则if（Char ==''U''）
返回''H'';
否则if（Char ==''V''）
返回''X'';
else if（Char ==''W''）
返回''C'';
否则if（Char ==''X''）
返回''V'';
else if（Char ==''Y''）
return' G'';
否则if（Char ==''Z''）
返回''R'';
否则如果（ Char ==''0''）
返回''A'';
否则if（Char ==''1''）
返回''9'';
否则if（Char ==''2''）
返回''8'';
else if（Char ==''3''）
返回''7'';
else if（Char ==''4''）
返回''6'';
else if（Char ==''5''）
返回''5'';
else if（Char ==''6''）
返回''4'';
else if（Char =='' 7''）
返回''3';
否则if（Char ==''8''）
return'' 2'';
否则if（Char ==''9''）
返回''1';
}
如果您需要更多信息，请询问，我会生产它。谢谢!!!!
I''ve written this program that simulates a 36 character, 10 rotor
reciprocal rotor cipher, w/ a plugboard. Any way I can make the
plugboard function more compact and/or be able to change the mapping
at runtime?
char Enigma::plugboard(char Char)
{
if(Char==''A'')
return ''0'';
else if(Char==''B'')
return ''Q'';
else if(Char==''C'')
return ''W'';
else if(Char==''D'')
return ''E'';
else if(Char==''E'')
return ''D'';
else if(Char==''F'')
return ''T'';
else if(Char==''G'')
return ''Y'';
else if(Char==''H'')
return ''U'';
else if(Char==''I'')
return ''I'';
else if(Char==''J'')
return ''O'';
else if(Char==''K'')
return ''P'';
else if(Char==''L'')
return ''S'';
else if(Char==''M'')
return ''N'';
else if(Char==''N'')
return ''M'';
else if(Char==''O'')
return ''J'';
else if(Char==''P'')
return ''K'';
else if(Char==''Q'')
return ''B'';
else if(Char==''R'')
return ''Z'';
else if(Char==''S'')
return ''L'';
else if(Char==''T'')
return ''F'';
else if(Char==''U'')
return ''H'';
else if(Char==''V'')
return ''X'';
else if(Char==''W'')
return ''C'';
else if(Char==''X'')
return ''V'';
else if(Char==''Y'')
return ''G'';
else if(Char==''Z'')
return ''R'';
else if(Char==''0'')
return ''A'';
else if(Char==''1'')
return ''9'';
else if(Char==''2'')
return ''8'';
else if(Char==''3'')
return ''7'';
else if(Char==''4'')
return ''6'';
else if(Char==''5'')
return ''5'';
else if(Char==''6'')
return ''4'';
else if(Char==''7'')
return ''3'';
else if(Char==''8'')
return ''2'';
else if(Char==''9'')
return ''1'';
}
If you need any more info, just ask and I''ll produce it. Thanks!!!!
推荐答案
Protoman写道：
Protoman wrote:
I 这个程序写的模拟了一个36个字符，10个转子
互易转子密码，带有一个插板。我能用什么方式使
插板功能更紧凑和/或能够在运行时更改映射
？
I''ve written this program that simulates a 36 character, 10 rotor
reciprocal rotor cipher, w/ a plugboard. Any way I can make the
plugboard function more compact and/or be able to change the mapping
at runtime?
在这种情况下，开关可能更紧凑：
开关（字符）{
case''A''：返回''0'';
case''B''：返回''Q'';
....
}
如果你想在运行时更改：
#include< map>
class Enigma {
std :: map< char，charplug;
public：
char plugboard（char Char）{
返回插头[Char];
}
....
};
-
rbh
In this situation, a switch might be more compact:
switch (Char) {
case ''A'': return ''0'';
case ''B'': return ''Q'';
....
}
If you want to change at runtime:
#include <map>
class Enigma {
std::map<char, charplug;
public:
char plugboard(char Char) {
return plug[Char];
}
....
};
--
rbh
7月2日20:10，Robert Bauck Hamar< roberth + n ... @ ifi .uio.nowrote：
On 2 Jul, 20:10, Robert Bauck Hamar <roberth+n...@ifi.uio.nowrote:
Protoman写道：
Protoman wrote:
我写了这个程序来模拟一个36个字符，10个转子
互易转子密码，带插板。我能用什么方式使
插板功能更紧凑和/或能够在运行时更改映射
？
I''ve written this program that simulates a 36 character, 10 rotor
reciprocal rotor cipher, w/ a plugboard. Any way I can make the
plugboard function more compact and/or be able to change the mapping
at runtime?
在这种情况下，开关可能更紧凑：
开关（字符）{
case''A''：返回''0'';
case''B''：返回''Q'';
...
}
如果你想在运行时更改：
#include< map>
class Enigma {
std :: map< char，charplug;
public：
char plugboard（char Char）{
返回插头[Char];
}
...
} ;
-
rbh
In this situation, a switch might be more compact:
switch (Char) {
case ''A'': return ''0'';
case ''B'': return ''Q'';
...
}
If you want to change at runtime:
#include <map>
class Enigma {
std::map<char, charplug;
public:
char plugboard(char Char) {
return plug[Char];
}
...
};
--
rbh
哦，加密（）没有''工作正常...它只返回明文的第一个
字符：
字符串Enigma :: Encrypt（const string& cleartext）
{
string ciphertext;
ciphertext.resize（cleartext.size（））;
unsigned int i = 0;
for（; i< cleartext.length（）; i ++）
{
int val = Rotor :: Ch aracterMap（plugboard（cleartext [i]））;
if（val< 36）
{
char val1 = R1.GetCharacterIndex（val ）;
int val2 = Rotor :: CharacterMap（val1）;
char val3 = R2.GetCharacterIndex（val2）;
int val4 = Rotor :: CharacterMap（val3）;
char val5 = R3.GetCharacterIndex（val4）;
int val6 = Rotor :: CharacterMap（val5）;
char val7 = R4.GetCharacterIndex（val6）;
int val8 = Rotor :: CharacterMap（val7）;
char val9 = R5.GetCharacterIndex（val8）;
int val10 = Rotor :: CharacterMap（val9）;
char val11 = R6.GetCharacterIndex（val10）;
int val12 =转子： ：CharacterMap（val11）;
char val13 = R7.GetCharacterIndex（val12）;
int val14 = Rotor :: CharacterMap（val13）;
char val15 = R8.GetCharacterIndex（val14）;
int val16 = Rotor :: CharacterMap（val15）;
char val17 = R9.GetCharacterIndex（val16）;
int val18 = Rotor :: CharacterMap（val17）;
char val19 = R10.GetCharacterIndex（val18）;
int val20 = Rotor :: CharacterMap（val19）;
char val21 = Enigma :: Reflector [val20];
int val22 = Rotor :: CharacterMap（val21）;
char val23 = R10.GetCharacterInverse（val22）;
int val24 = Rotor :: CharacterMap（val23 ）;
char val25 = R9.GetCharacterInverse（val24）;
int val26 = Rotor :: CharacterMap（val25）;
char val27 = R8.GetCharacterInverse（val26）;
int val28 = Rotor :: CharacterMap（val27）;
char val29 = R7.GetCharacterInverse（val28）;
int val30 = Rotor :: CharacterMap（val29）;
char val31 = R6.GetCharacterInverse（val30）;
int val32 = Rotor :: CharacterMap（val31）;
char val33 = R5.GetCharacterIndex（val32）;
int val34 = Rotor :: CharacterMap（val33）;
char val35 = R4。 GetCharacterIndex（val34）;
int val36 = Rotor :: CharacterMap（val35）;
char val37 = R3.GetCharacterIndex（val36）;
int val38 = Rotor :: Charac terMap（val37）;
char val39 = R2.GetCharacterIndex（val38）;
int val40 = Rotor :: CharacterMap（val39）;
char val41 = R1.GetCharacterIndex（val40）;
ciphertext [i] = plugboard（val41）;
R1.AdvanceRotor（1）;
if（（R1.GetSteps（）％36）== 0）
{
R2.AdvanceRotor（1）;
if（ （R2.GetSteps（）％36）== 0）
{
R3.AdvanceRotor（1）;
if（（R3。 GetSteps（）％36）== 0）
{
R4.AdvanceRotor（1）;
if（（R4.GetSteps（） ％36）== 0）
R5.AdvanceRotor（1）;
{
if（（R5.GetSteps（）％36） == 0）
R6.AdvanceRotor（1）;
{
if（（R6.GetSteps（）％36）== 0 ）
R7.AdvanceRotor（1）;
{
if（（R7.GetSteps（）％36）== 0）
R8.AdvanceRotor（1）;
{
if（（R8.GetSteps（）％36）== 0）
R9.AdvanceRotor（1）;
{
if（（R9.GetSteps（）％36）== 0）
R10.AdvanceRotor（1）;
}
}
}
}
}
}
}
} //问题＃2缺少大括号
else {ciphertext [i] = cleartext [i];}
}
返回密文;
}
}
Oh, and the encrypt() doesn''t work right...it just returns the first
char of the cleartext:
string Enigma::Encrypt(const string& cleartext)
{
string ciphertext;
ciphertext.resize(cleartext.size());
unsigned int i=0;
for(;i<cleartext.length();i++)
{
int val=Rotor::CharacterMap(plugboard(cleartext[i]));
if (val<36)
{
char val1 = R1.GetCharacterIndex(val);
int val2 = Rotor::CharacterMap(val1);
char val3 = R2.GetCharacterIndex(val2);
int val4 = Rotor::CharacterMap(val3);
char val5 = R3.GetCharacterIndex(val4);
int val6 = Rotor::CharacterMap(val5);
char val7 = R4.GetCharacterIndex(val6);
int val8 = Rotor::CharacterMap(val7);
char val9=R5.GetCharacterIndex(val8);
int val10 = Rotor::CharacterMap(val9);
char val11=R6.GetCharacterIndex(val10);
int val12=Rotor::CharacterMap(val11);
char val13=R7.GetCharacterIndex(val12);
int val14=Rotor::CharacterMap(val13);
char val15=R8.GetCharacterIndex(val14);
int val16=Rotor::CharacterMap(val15);
char val17=R9.GetCharacterIndex(val16);
int val18=Rotor::CharacterMap(val17);
char val19=R10.GetCharacterIndex(val18);
int val20=Rotor::CharacterMap(val19);
char val21 = Enigma::Reflector[val20];
int val22 = Rotor::CharacterMap(val21);
char val23 = R10.GetCharacterInverse(val22);
int val24 = Rotor::CharacterMap(val23);
char val25 = R9.GetCharacterInverse(val24);
int val26 = Rotor::CharacterMap(val25);
char val27 = R8.GetCharacterInverse(val26);
int val28 = Rotor::CharacterMap(val27);
char val29 = R7.GetCharacterInverse(val28);
int val30 = Rotor::CharacterMap(val29);
char val31 = R6.GetCharacterInverse(val30);
int val32=Rotor::CharacterMap(val31);
char val33=R5.GetCharacterIndex(val32);
int val34=Rotor::CharacterMap(val33);
char val35=R4.GetCharacterIndex(val34);
int val36=Rotor::CharacterMap(val35);
char val37=R3.GetCharacterIndex(val36);
int val38=Rotor::CharacterMap(val37);
char val39=R2.GetCharacterIndex(val38);
int val40=Rotor::CharacterMap(val39);
char val41=R1.GetCharacterIndex(val40);
ciphertext[i] = plugboard(val41);
R1.AdvanceRotor(1);
if((R1.GetSteps()%36)==0)
{
R2.AdvanceRotor(1);
if((R2.GetSteps()%36)==0)
{
R3.AdvanceRotor(1);
if((R3.GetSteps()%36)==0)
{
R4.AdvanceRotor(1);
if((R4.GetSteps()%36)==0)
R5.AdvanceRotor(1);
{
if((R5.GetSteps()%36)==0)
R6.AdvanceRotor(1);
{
if((R6.GetSteps()%36)==0)
R7.AdvanceRotor(1);
{
if((R7.GetSteps()%36)==0)
R8.AdvanceRotor(1);
{
if((R8.GetSteps()%36)==0)
R9.AdvanceRotor(1);
{
if((R9.GetSteps()%36)==0)
R10.AdvanceRotor(1);
}
}
}
}
}
}
}
} // problem #2 missing brace
else {ciphertext[i] = cleartext[i];}
}
return ciphertext;
}
}
Protoman写道：
Protoman wrote:
我''我写了这个程序模拟一个36个字符，10个转子
互易转子密码，带有一个插板。我有什么方法可以使
插板功能更紧凑和/或能够在运行时更改映射
？
char Enigma :: plugboard（char char）
{
if（Char ==''A''）
返回''0' ';
否则if（Char ==''B''）
返回''Q'';
I''ve written this program that simulates a 36 character, 10 rotor
reciprocal rotor cipher, w/ a plugboard. Any way I can make the
plugboard function more compact and/or be able to change the mapping
at runtime?
char Enigma::plugboard(char Char)
{
if(Char==''A'')
return ''0'';
else if(Char==''B'')
return ''Q'';
....
将所有输出放入char数组[] = {''0''， ''Q''，...};
返回数组[Char - ''A''];
在运行时更改映射你可以把文件读进阵列。
-
Scott McPhillips [MVP VC ++]
....
Put all the outputs in a char array[] = {''0'', ''Q'', ...};
return array[Char - ''A''];
To change the mapping at run time you could read a file into the array.
--
Scott McPhillips [MVP VC++]
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