String with String.fromCharCode ()

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1

The following script found here in the Stack prints from "aaa" to "zzz" by incrementing letters in order, one by one: 

var str= 'aaa',
s= str;

while(str!=='zzz') {
  str= ((parseInt(str, 36)+1).toString(36)).replace(/0/g,'a');
  s+= '<br/> '+str;
}

document.body.innerHTML= s;

Output: 

aaa
aab
aac
...
aba
abb
abc
...
zzy
zzz

My intention is to continue this logic, but including special characters, using String.fromCharCode() of index 33 to 126.

This numeric range already includes all special characters, letters, and numbers.

  

The desired output should contain 16 characters, not just the three   above.

What modifications would be needed in the script to accomplish this task?

    
asked by anonymous 17.06.2016 / 07:17

3 answers

1

In theory the code will solve your problem, I used the variable "n" as a pointer to know which position of the array should increase, but I believe that in practice it will take a lot of time to execute this script regardless of the medium being used. / p> 

var v = [33,33,33,33,33,33,33,33,33,33,33,33,33,33,33,33];
var result = [];
var n = 15;
result.push(String.fromCharCode(v[0],v[1],v[2],v[3],v[4],v[5],v[6],v[7],v[8],v[9],v[10],v[11],v[12],v[13],v[14],v[15]));

while (v[n] != 126 || n != 0) {
  if (v[n] == 126 && n != 0){
    v[n] = 33;
    n--;
  }
  else {
    if (v[n] != 126) {
      v[n]++;
      result.push(String.fromCharCode(v[0],v[1],v[2],v[3],v[4],v[5],v[6],v[7],v[8],v[9],v[10],v[11],v[12],v[13],v[14],v[15]));
      n = 15;
    }
    else {
      result.push(String.fromCharCode(v[0],v[1],v[2],v[3],v[4],v[5],v[6],v[7],v[8],v[9],v[10],v[11],v[12],v[13],v[14],v[15]));
    }
  }
};

var retorno = result.join('<br/>');
    
18.06.2016 / 04:50
1

The simplest way in this case is to do three cycles for chained, but due to the large number of results (1 092 727) it will become a little slow ... in the test I did took 28,75s 

var str='';
s= str;

for(i=33;i<=126;i++)
  for(j=33;j<=126;j++)
    for(k=33;k<=126;k++)
      str+=String.fromCharCode(i)+String.fromCharCode(j)+String.fromCharCode(k)+' <br/> ';

document.body.innerHTML= str;

Output: 

!!!
!!"
!!# 
...
"<@|
<@}
<@~
"
    
17.06.2016 / 13:23
1

To do this you need, I recommend using "String.fromCharCode".

It would look something like this: 

var s = [];
var retorno = '';

for (c3 = 33; c3 < 126; c3++)
   for (c2 = 33; c2 < 126; c2++)
      for (c1 = 33; c1 < 126; c1++)
         s.push(String.fromCharCode(c1,c2,c3));
      
retorno = s.join('<br/>');

I hope I have helped.

    
17.06.2016 / 13:38
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