Not Working for some reason

[valltax]

I'm new to c++ but i just try to make a simple text based game and the cin command wasnt working for the second time i was using it here is a chuck of my code.



#include <iostream>

using namespace std;
int sticks = 0;
int health = 100;//players health
int rocks = 0;
int pickaxehealth = 50;
int swordhealth = 50;
char name;


int main()
{
int direction1;
cout << "Welcome to Classic Text 1.0." << endl;
cout << "____________________________" << endl;
cout << " Enter your name: ";
cin >> name;
cout << "You wake up in the middle of the night, In a" << endl <<
"clearing of a forest. There is a path to the" << endl <<
"left and the right. Which way do you want to go" << endl;
cout << "_______________________________________________" << endl;
cout << " Type 1 to go right. Type 2 to go left"<< endl;
cout << "Enter which way you want to go: " << endl;
cin >> direction1;

[Zhuge]

What do you mean by "it wasn't working"?

[jim hurley]

Try char* name; instead of char name;. You're using a single character instead of a character array.

(Or at least I think that's the problem. Zhuge, correct me if I'm wrong.)

[valltax]

no thats working but cin >> direction1; isnt

[valltax]

every thing works zhuge but it doesn't allow the user to type in direction1. it shut down the program with out asking you what you what to import for direction1.

[jim hurley]

Even if it's not the problem you are currently having, you should change it still because your variable name is only withholding the first character you input.

http://www.cplusplus.com/forum/articles/6046/

This may be addressing your other problem.

[valltax]

well right now i am reading c++ primer 5th edition, so it will probably show up in there soon.

[tmihos]

You send a stream (std::cin) to a char pointer (for one or more chars but pointer) Stream need a buffer.
Declare name as: char name[32]. Now if std::cin data are less than 32 std::cin frees its space and send new data.
In your program std::cin access to direction but still have char data because name holds only the 1st byte.