c++ Replacing character with a number [closed] - c++

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I have this code. What i wanted to do is to replace every letter from every char with indicated number. Like for A is 10, and so on until J, 19. My code works well if i have only one letter in my char array, but if i have more after another it copies useless things. I think that something is wrong with strncat.
#include<conio.h>
#include <cstring>
#include <iostream>
using namespace std;
int main()
{
char litera[11]={"ABCDEFGHIJ"};
char cifra[11]={"0123456789"};
char rezultat[256]={0};
int n;cin>>n;cin.get();
for(int i=0;i<n;i++)
{
char x[256];
cin.get(x,256);cin.get();
int k=strlen(x);
for(int j=0;j<k;j++)
{
int p = strchr(litera,x[j])-litera;
if(p>=0 )
{
strncat(rezultat, x,j);
// my p is the indicator for letter and number
strcat(rezultat,"1");
// I'm copying an 1 there because i have numbers
// between 10-19 and in my int array i have only
// the second cipher
strcpy(x,x+j);
rezultat[strlen(rezultat)]=cifra[p];
}
}
cout<<rezultat<<endl;
memset(rezultat,0,sizeof(rezultat));
}
getch();
return 0;
}
Input: 07B, 1AA, C8A
Output: 0711, 11010, 12810
My output:
0711
110AA1
12C810
If you guys can tell me where the problem is, you'll help me a lot, every suggestion is well received, even if is not about this problem. Thanks!

Answer 1

If it is allowed to use vector and string then you can try the following (I tested it and it works.) The input here should be line by line(each line is a new string to be converted) but of course you may modify this part based on your input format:
Include those:
#include <vector>
#include <iostream>
#include <string>
//#include <conio.h>
//#include <cstring>
//using namespace std;
int main()
{
int n;
std::cin >> n;
while (n--)
{
string input;
std::cin >> input;
vector<char> resultat;
for (int i = 0; i < input.size(); i++)
{
if (input.at(i) >= 'A' && input.at(i) <= 'J') // it is a letter between A and J
{
int number = input.at(i) - 'A' + 10;
char ones = (char)(((int)'0') + (number % 10));
char tens = (char)(((int)'0') + (number / 10));
resultat.push_back(tens);
resultat.push_back(ones);
}
else
resultat.push_back(input.at(i));
}
for (int i = 0; i < resultat.size(); i++)
{
std::cout << resultat[i];
}
std::cout << endl;
}
}
if string is not allowed just use your character array. There seems to be nothing wrong in your input format.
If vector is not allowed either, then you may create a char pointer the initialize it based on the final size.
For example:
int newSize=0;
for (int i = 0; i < input.size(); i++)
{
if (input.at(i) >= 'A' && input.at(i) <= 'J')
newSize+=2;
else
newSize++;
}
char* resultat = new char[newSize];
...
Then just fill the resultat.
Hope that helps!

Answer 2

Do you have to do it with your multiple arrays? You could perform a switch-case for only specified inputs. Since you are replacing letters A-J with numbers 10-19, you can easily implement this in switch case. You can also check for incorrect input as well.
If you have to use arrays, first question is why is you numerical array the single digits? Is this is constraint? Cant you use :
char cifra[11]={"10","11","12","13","14","15","16","17","18","19"};
for the array instead, that way you can refer to the array index and replace it using the index? This way, you can compliment the arrays by just using index referencing to replace the output based on the index, such as A is index 1 for alphabetical array which refers to index 1 of the numerical array which is "10" so you just print the array index value to a variable and output it.