Checking to see if integer is a character.

[thiskid]

#include <iostream>
using namespace std;

int main()
{
int x;

cout << "Enter character:";
cin >> x;
if(char(x)=='+')
{
cout << "This is addition";
}
else
{
cout << "This is not addition";
}
return 0;
}

If i type in:

+

How come it says that "this is not addition?

[S G H]

Change from
int x;
to
char x;

This will read to the variable "x" in "Character mode" (As text) and not as a number.
Using int will read to "x" as "Number mode".

(Terminology's made up, for simplicity's sake)

[Guzfraba]

char(x) returns the character with the ASCII code number x. So, just if you enter 43, which is the ASCII code for '+', you will get "This is addition".
You cannot store characters in int data type, just ASCII code representing characters. Use char data type for character handling.

[thiskid]

I want to keep the data type int. But I want to write a program that determines whether the user did enter a plus sign instead of a regular number. Is there a way to check to see if a data type int was entered as an addition sign?

[Guzfraba]

I don't think so,entering a non-number value instead of a integer, will cause cin to fail. Instead, you can use char data type and check if character is number,or if it's non-number.
http://www.cplusplus.com/reference/cctype/

[JLBorges]

The slightly tricky part is that +6 is valid input of a number.

Perhaps, something like this:

#include <iostream>

int main()
{
    int x ;
    char c ;
    bool number = false ;

    std::cout << "Enter either a number or a character:";

    std::cin >> c ; // read a char

    // put it back and try to read a number
    std::cin.putback(c) ;
    if( std::cin >> x ) number = true ;
    else
    {
        std::cin.clear() ; // clear the failed state (to enable more input)
        x = c ;
    }

    if(number) std::cout << "This is the number " << x << '\n' ;

    else if( x == '+' ) std::cout << "This is addition\n" ;

    else std::cout << "This is something else\n" ;
}