Compile Error -> Integer to Char
Posted: Wed Sep 01, 2004 1:30 am
Greetings!
Need some help here
How can I convert an Integer into a Char?
I am getting a "Compile Error" in one program.
The message includes:
[cpp]#include <iostream.h>
#include <stdlib.h>
...
char Minesweep()
{ unsigned short Mines = 0;
char AuxC[1];
if (Matriz[i-1][j] == '*')
Mines++;
if (Matriz[i-1][j+1] == '*')
Mines++;
if (Matriz[j+1] == '*')
Mines++;
if (Matriz[i+1][j+1] == '*')
Mines++;
if (Matriz[i+1][j] == '*')
Mines++;
if (Matriz[i+1][j-1] == '*')
Mines++;
if (Matriz[j-1] == '*')
Mines++;
if (Matriz[i-1][j-1] == '*')
Mines++;
itoa(Mines,AuxC,10);
return AuxC[0];
}
...
[/cpp]
It runs just fine in Dev-C++ 4.
What can I do?
Thanks in advance!
Need some help here

How can I convert an Integer into a Char?
I am getting a "Compile Error" in one program.
The message includes:
In my code I have:In function `char Minesweep()':
implicit declaration of function `int itoa(...)'
[cpp]#include <iostream.h>
#include <stdlib.h>
...
char Minesweep()
{ unsigned short Mines = 0;
char AuxC[1];
if (Matriz[i-1][j] == '*')
Mines++;
if (Matriz[i-1][j+1] == '*')
Mines++;
if (Matriz[j+1] == '*')
Mines++;
if (Matriz[i+1][j+1] == '*')
Mines++;
if (Matriz[i+1][j] == '*')
Mines++;
if (Matriz[i+1][j-1] == '*')
Mines++;
if (Matriz[j-1] == '*')
Mines++;
if (Matriz[i-1][j-1] == '*')
Mines++;
itoa(Mines,AuxC,10);
return AuxC[0];
}
...
[/cpp]
It runs just fine in Dev-C++ 4.
What can I do?
Thanks in advance!