[C++] Why won't this compile?

[2GetherWeRise]

So those of you who saw my topic yesterday told me that the variable type "char" can only store one letter, as opposed to a word. So i made a new script using letters instead of words for the variable, yet my compiler still says there is an error. Can someone tell me what the error is?

#include <iostream>

using namespace std;

int main()
{
	char choice;

	cout<<"Welcome to the Super Quiz Game!\n";
	cout<<"I will be asking you a series of\n";
	cout<<"of multiple choice questions, and\n";
	cout<<"see how many you can answer! Ready?\n";
	cin.get();
	cout<<"lets begin\n";
	cin.get();
	cout<<"What two nations fought in the War of 1812?\n";
	cout<<"A. France and Canada\n";
	cout<<"B. America and Britian\n";
	cout<<"C. Mexico and America\n";
	cin>> choice;
	cin.ignore();

	if ( choice == b ) {
		cout<<"You are correct!\n";
	}
	cin.get();
}

Thank you to all who answer.

[MentaL]

[IQstim]

quote tags my man. qoute tags.

choice == b
choice == "b"

BAM BAM BAM! you're dead.

[2GetherWeRise]

i put quote tags and now the compiler says theres an error in the "==" and it says operand types are incompatible.

[Generic230]

Limited knowledge:
You use '' instead of "" for chars. Speaking without any C++ knowledge, but in Java you use '' for a char

[2GetherWeRise]

Uh, run that by me again? i don't really understand what you just said.

[Negata]

A single character is written like 'b'. If you type "b", it would mean it's a string with just one letter, and behind the curtains the actual data is two bytes, one for the character b and one 0, as C-strings need the 0 for marking where the string ends in memory. If there wasn't such a marker, there would be no way for the program to know how long the actual string is. Again in your listing (choice == b) is interpreted like both choice and b are variable names.

So, obviously you can't store "b", being 2 bytes, to a char that only holds one.

[Daevius]

quote tags my man. qoute tags.

choice == b
choice == "b"

BAM BAM BAM! you're dead.

BAM BAM BAM! you're dead too!

"b" is a pointer to an array of characters in the stack
Checking if it's equal to 'b' would be the right thing to do here, the input is also put in a characterm so that's valid.

Btw, to the OP; did you not get a compiler warning saying the variable b is never declared? Or did you ignore it / did not understand it?
I also suggest leaving out the 'using' line, and prefix the standard lib with 'std::', but that's me.

Also, you get a warning for not returning an integer. End main with 'return 0;'.