error while displaying array of pointers to pointers

[Goofy]

Why can't I display the array of pointers in the 'display' function, when it worked in the 'position' function,


\****************MAIN********************\

#include <iostream>
using namespace std;

#include "Table.h"

int main()
{
Table *ptrtoF;
int length, width;
char fig1;
cout << "Give any charachter: "<<endl;
cin>>fig1;
cout << "\n0-7 for the position on the table ";
cin>> length ;
cout << "\n0-7 for the position on the table : ";
cin>> width;
ptrtoF->position(fig1,length, width);
ptrtoF->display();

system("pause");
return 0;
}



\************ HEADER FILE**************\
#include <iostream>
using namespace std;
char **table[8][8];
char character='X';
class Table
{

public:
void position(char s, int a , int b)
{

char * temp;
char * temp1;
temp=&character;

cout <<endl<<*temp<<endl;

for(int i=0;i!=8;i++)
{for(int s=0;s!=8;s++)
{
table[i][s]=&temp;}}

temp1=&s;
table[a][b] =&temp1;

for(int i=0;i!=8;i++)
{for(int s=0;s!=8;s++)
{cout << **table[i][s]<<(s==7?"\n":" ") ;}}} //here it works

int display()
{

for(int i=0;i!=8;i++)
{for(int s=0;s!=8;s++)
cout <<"here it stops working: ERROR"<<**table[i][s]<<(s==7?"\n":" ");}} // here it doesn't

};



\****\
it should be a table where you put some char in the position you want , and than displays it when you ask for, but it only works in the position function. I know it can be Simpler ,even not using pointers at all , but I need it for some other thing I'm working at and naturally and most importantly I want to learn it for next time.

I really appreciate your generous help, which cplusplus members every time offer.

[spaggy]

uhm not sure if this is the issue but do you notice that your position function takes in a char s but also declares an int s in its body. temp1=&s; <-- which s do you think this is using char s or int s. If it uses int s the variable temp1 is pointing to gets deleted once you exit the function scope which will cause errors in display()

[guestgulkan]

In the position function

	void position(char s, int a , int b)
	{

		char * temp;//local variable
		char * temp1;
		temp=&character;

		cout <<endl<<*temp<<endl;

		for(int i=0;i!=8;i++)
		{
			for(int s=0;s!=8;s++)
			{
				table[i][s]=&temp;//error  - assigning address of local variable
			
			}
		}

You are filling up the the table with the address of a variable (temp) which is
is local to the position function - this is why it all works when inside the position function, but
the variable (temp) will be destroyed when the position function finishes so it's address is no longer valid.

[Goofy]

Thanks spaggy and guestgulkan it works yet i declared them global and called 's' by reference, works tip top.

void position(char &s, int a , int b)