Beginner - calculator

[MrLazio]

Hello. I am just starting C programmer if I can call myself a programmer right now but let's get straight into the problem. The main point of the program is to calculate equations just like a standard calculator but I wanted to do it myself. I don't understand what the problem is right now but I've managed to create a program that asks for both values but somehow it doesn't want to ask for an operator (*, /, + etc). What's wrong with my code that the terminal skips the scanning part for the operator?

#include <stdio.h>

int main()
{
	int value1, value2, answer;
	char operator;
	
	printf("Enter the first value: ");
	scanf("%i", &value1);
	
	printf("Enter the second value: ");
	scanf("%i", &value2);
	
	printf("Enter the operator: ");
	scanf("%c", &operator);
	
	if(operator == "+")
		answer = value1 + value2;
	if(operator == "-")
		answer = value1 - value2;
	if(operator == "*")
		answer = value1 * value2;
	if(operator == "/")
		answer = value1 / value2;
	
	printf("This is your result:\n");
	printf("%i\n", answer);
	
	return 0;
}

[Matticus]

Welcome to the forum. This entry in our FAQ has an explanation for the problem you're seeing: FAQ > How do I avoid a "dangling" newline when reading single character user input? - Cprogramming.com
If you need clarification, just ask.

[MrLazio]

Okay, I get what the problem is now. I've changed it but why does the answer always becomes 0? I don't see any problem with the if statement. I've told the program to add value1 and value2 if the operator entered is "+". So, what is the actual problem?

[tabstop]

The answer isn't always zero. But zero is definitely a possibility; for instance 1/2 is zero (well, it's zero remainder one, but you didn't ask for the remainder).

[MrLazio]

Wait, I don't get it. So how should it look then?

[tabstop]

I suppose the question is what kind of output do you want to get?

[MrLazio]

Well the output of a standard calculator. If the first value is 5, second value is 5 and an operator is + then the answer should be equalled to 10. That's the point I'm trying to make.

[tabstop]

Well the output of a standard calculator. If the first value is 5, second value is 5 and an operator is + then the answer should be equalled to 10. That's the point I'm trying to make.

And that's what you have. (Assuming you actually did fix the read-the-operator issue. If you didn't fix the read-the-operator issue, well then, you would need to fix that.)

[MrLazio]

Look at the screenshot below. The coding looks OK. It reads the operator now but doesn't give the expected answer. While it should be 10 it still gives 0 in any case. Rather it is 5*5 or 10/2, it is always 0. Have a look.
http://iv.pl/images/71653172978004629313.png

[Matticus]

For a single character, you should only be using single quotes.

// incorrect
if(operator == "+")

// correct
if(operator == '+')

[MrLazio]

For a single character, you should only be using single quotes.

You're the beast Matticus! Thank you!

[tabstop]

For a single character, you should only be using single quotes.

// incorrect
if(operator == "+")

// correct
if(operator == '+')

Good catch! (The Nyquil is kicking in and I haven't even taken it yet! Sigh.)

Your compiler should be screaming at you about that; mine says "warning: comparison between pointer and integer" for each of those lines.

[Matticus]

You're the beast Matticus! Thank you!

I'll take that as a compliment
(Roar)