CareerCup

#include <string>
#include <iostream>
#include <sstream>
#include <iomanip>

using namespace std;

string convert(const string &s){
	stringstream o;

  	for(char i : s){
    		if(i < ' ')
      			o << "\\" << setw(3) << setfill('0') << oct << (int)i;
    		else
     			 o << i;
  	}
  	return o.str();
}

int main(int argc, char *argv[]) {
  	cout << convert("A\tB") << endl;
  	return 0;
}

output : A\011B

Use a lookup table for the 32 special ASCII characters and then simply search and replace.


public class OctalEscape {

public static void main(String args[]) {
System.out.println(escape("\000sdjfj k\004ljf lafj \100\040\100 asdfj akl\010jsdfklf"));
}

static String lookup[];
static {
buildLookupTable();
}

static void buildLookupTable() {
lookup = new String['0'];
char octal[] = new char[] {'0','0','0'};
int c = 0;
for (char i = '0'; i <= '7'; ++i) {
octal[0] = i;
for (char j = '0'; j <= '7'; ++j) {
octal[1] = j;
for (char k = '0'; k <= '7'; ++k) {
octal[2] = k;
lookup[c++] = "\\" + octal[0] + octal[1] + octal[2];
if (c == ' ') {
return;
}
}
}
}
}

static String escape(String s) {
StringBuffer result = new StringBuffer();
char sArray[] = s.toCharArray();
for (char c : sArray) {
if (c < ' ') {
result.append(lookup[c]);
}
else {
result.append(c);
}
}
return result.toString();
}
}

Hi

Can you please elaborate ?

Hi, basically, nothing should be modified for the original string, since the new char '\ooo' would be the same as the previous char.

So if I understand you correctly if the input string is: "A#B" output should be "A\043B" as ascii character '#' maps to decimal 35 and octodecimal 43. Right?