cant figure it out

[Dark Ares]

I need help to understand this problem.

#include <iostream>

using namespace std;

int main ()
{
    int password;
    
    cout<<"Password: ";
    cin>> password;
    cin.ignore();
    if (password == 100) {
            cout<<"Access cleared\n";
            }
    else {
         cout<<"Access denied\n";
         }
    cin.get();
}

[Dark Ares]

how come the compiler stops when i put david for 100?
how do i put an alfabeth password? do i need char?

[Dark Ares]

must i put char at
int password?

[adr]

you could use a char or a string for alfabeth password.
string password; //<~~string form

Also it stops b/c the int is looking for a interger not a word or letter unlike a char or string.

[Salem]

> how come the compiler stops when i put david for 100?
Be careful of your terminology, when you run the program you're no longer anywhere near the compiler.

The compiler turns your source code into an executable.
You then run your executable and enter input.

[Dark Ares]

> how come the compiler stops when i put david for 100?
Be careful of your terminology, when you run the program you're no longer anywhere near the compiler.

the compiler stops saying something about an error and it makes the line with the 100 [or david] red.

How would the code be then?

[Salem]

Well post your error messages then!
Or try www.psychic-programming-network.com

[Dark Ares]

here look.

[cloudy]

Right 2 things wrong with it.

if(password==david)

The compiler thinks that david is a variable that you havent declared. You are trying to use it as a literal. You need to put speech marks round david like so:

if(password=="david")


second of all you have declared password as an integer and it needs to be of type char*.

change:
int password;
to
char* password;

[Dark Ares]

i thought so about the char. What about the asterisk?

[hk_mp5kpdw]

second of all you have declared password as an integer and it needs to be of type char*.

change:
int password;
to
char* password;

What it needs to be (in order to be used in the context you are suggesting, i.e. using the equality operator == to compare with a string literal) is a string and not an uninitialized pointer pointing to god knows where.

i thought so about the char. What about the asterisk?

Ignore that for now (it means that the variable is a pointer), just use a string.

#include <iostream>
#include <string>

using namespace std;

int main ()
{
    string password;
    
    cout<<"Password: ";
    cin>> password;
    cin.ignore();
    if (password == "david") {
            cout<<"Access cleared\n";
            }
    else {
         cout<<"Access denied\n";
         }
    cin.get();
}

[cloudy]

you dont have to use the asterix, if you wanted to you could do

char password[20];

you can replace the number 20 with what ever integer you want, but the password cannot exceed that length.

This is some basic stuff, do you have any c books because the first 2 chapters can help you with this stuff mate.

[twomers]

try:

#include <iostream>
#include <string>
using namespace std;

int main ()
{
    string password;
    
    cout<<"Password: ";
    cin >> password;

    if (password == "David") 
	{
		cout<<"Access cleared\n";
	}
    else
	{
		cout<<"Access denied\n"; 
	}
	return 0;
}

This will also work for numbers:

#include <iostream>
#include <string>
using namespace std;

int main ()
{
    string password;
    
    cout<<"Password: ";
    cin >> password;

    if (password == "42") 
	{
		cout<<"Access cleared\n";
	}
    else
	{
		cout<<"Access denied\n"; 
	}
	return 0;
}

I think 42 was an appropriate choice don't you? May not be the untimate answer, but it's pretty good.

[Dark Ares]

i just started an hour ago with the c++ tutorial on this site and whanted to try my learnings. the 20 in the char password[20] is the max lenght?
i will try now

[twomers]

the 20 in the char password[20] is the max lenght?

Yeah, it is. However it's easier if you #include <string> cause you don't have to declare its length. You can declare strings by going:

string mystring1 = "Place ";
string mystring2 = "text ";
string mystring3 = "here!";
string mystring4 = mystring1 + mystring2 + mystring3;

Output:

Place text here!

But if you're used to doing the char[len], it's just as good. You have to remember that if you want a string with 10 characters in it, you must allow for 11, this is because the last element of any string must be the terminating character or something.