convert int to char without itoa() [closed] - c++

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i have a program and i should convert int to char into it
but i cant use itoa() because site's judge don't support it
so i wrote this:
xt[i]= rmn+'0';
but i get this error :
Runtime error: Illegal file open (/dev/tty)
How should i convert this?
My code is this : (for palsquare question of USACO)
/*
ID: sa.13781
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <fstream>
using namespace std;
ofstream fout("palsquare.out");
int tool(char xt[])//CORRECT
{
int p = 0;
while (xt[p] != 0)
p++;
return p;
}
void prt(char xt[])//CORRECT
{
int p = 0;
while (xt[p] != 0)
{
fout << xt[p];
p++;
}
}
void mabna(int a, char xt[], int mab)
{
int ex = 1, tavan = 0, rmn, n;
for (; ex <= a; ex *= mab)
tavan++;
for (int i = tavan - 1; a != 0; i--, a /= mab)
{
rmn = a % mab;
switch (rmn)
{
case 0:
case 1:
case 2:
case 3:
case 4:
case 5:
case 6:
case 7:
case 8:
case 9:
xt[i] = rmn + '0';
break;
case 10:
xt[i] = 'A';
break;
case 11:
xt[i] = 'B';
break;
case 12:
xt[i] = 'C';
break;
case 13:
xt[i] = 'D';
break;
case 14:
xt[i] = 'E';
break;
case 15:
xt[i] = 'F';
break;
case 16:
xt[i] = 'G';
break;
case 17:
xt[i] = 'H';
break;
case 18:
xt[i] = 'I';
break;
case 19:
xt[i] = 'J';
break;
}
}
}
bool mirror(char *xt)//CORRECT
{
int p = 0;
int n = tool(xt);
for (int i = 0; i < (n / 2); i++)
if (xt[i] == xt[n - i - 1])
p++;
if (p == (n / 2))
return true;
return false;
}
void calc(int mab) //CORRECT
{
for (int i = 1; i <= 300; i++)
{
char p[10] = {0}, p2[10] = {0};
mabna(i * i, p2, mab);
if ( mirror(p2) == true )
{
mabna(i, p, mab);
prt(p);
fout << " ";
prt(p2);
fout << "\n";
}
}
}
int main()
{
ifstream fin("palsquare.in");
int mab;
fin >> mab;
calc(mab);
return 0;
}

Answer 1

Can you use sprintf()?
char s[16];
int x = 15;
sprintf(s, "%d", x);

Answer 2

If your int value is single digit, you can just cast it with (char)
If your int value is more than one digit, you can never expect a single char to hold it. char can only hold single character.
int num = 7;
char ch = (num + 48); //same effect as + '0'
cout << ch << endl;
OUTPUT: 7
If your int value is more than single digit. You need char* or char[] to hold the converted data. For example:
int num = 123987;
int len = 6;
int idx = len-1;;
char ch[20];
while(num > 0)
{
ch[idx] = (num%10) + 48;
num /= 10;
idx --;
}
for(int x=0; x<len; x++)
cout << "OUTPUT in char: " << ch[x];
OUTPUT in char: 123987
This is not a perfectly good way to handle this kind of situation, but it may gives you what you asked for.