returning an array using pointer

[olredixsis]

#include <iostream>

using namespace std;

char * userInput()

int main()
{
int i;
char filterInput[80];
char *(getInput)[80] = userInput();
cout<<"0: "<<getInput[4]<<endl;

for(i=0;i<80;i++){
if(getInput[i] == '\0'){
break;
}//end if
else{
filterInput[i] = getInput[i];
cout<<"filterInput"<<i<<" "<<filterInput[i]<<endl;
}
}//end for
cout<<"sd: "<<filterInput<<endl;
// braceCheck(filterInput[]);
cin.get();
cin.get();
cin.get();
cin.get();

}//end main

char * userInput()
{

char checkInput[80];
cout<<"Input Anything: ";
cin>>checkInput;
if((checkInput[0]=='e')&&(checkInput[1]=='x')&&(checkInput[2]=='i')&&(checkInput[3]=='t')){
exit (0); //terminates the program
}//end if
else{
char * character = checkInput;
return (character);
}//end else
}//end userInput function


-----------------------------------------------------------------------
what is the proper way of returning an array using a pointer?
I need to pass the array in userInput function to main.
How? I'm still confused with pointers to arrays thx.
Can someone edit my code to make it run properly? thx.

[Disch]

what is the proper way of returning an array using a pointer?

Generally, there isn't one. You don't return arrays.

You're better off passing a pointer to the buffer to the function:

void userInput(char* checkInput)
{
  // do stuff here
  // write to checkInput as if it were a normal array
}

int main()
{
  char input[80];

  // get user input
  userInput( input );

  // stuff here
}

Although -- screw char arrays. Save yourself the headaches and buffer overflows and use std::string:

string userInput()
{
  string checkInput;

  // do stuff with checkInput here

  return checkInput;
}

int main()
{
  string input = userInput;

  // etc
}

[olredixsis]

whoa! thank you very much Disch.
I never thought about that!