I just started teaching myself C++ on the Mac, and I have run into some issues.
I have written some code that allows the user to enter a number and when they hit enter, the number will be returned to the user.
Xcode will absolutely not have it though. Every time I try to run my code, it says that there is an issue with the cin>> thisisanumber; code. 
The error comes up and says
Invalid operands to binary expression. Error is on line 10.
What am I doing wrong?
#include <iostream>
using namespace std;
int main()
{
   int thisisanumber();
   cout << "Please enter a number: ";
   cin  >> thisisanumber;
   cin.ignore();
   cout << "You entered"<< thisisanumber <<"\n";
   cin.get();
}
                You've fallen victim to the most vexing parse, which means thisisanumber is being treated as a function. Take out the parentheses and you should be fine:
int thisisanumber;
Also consider making it a bit more readable, such as thisIsANumber. If you ever need to know it, thisIsANumber uses the camel-case naming convention.
Declare your variable without brackets, like
int thisisanumber;
With brackets, it is interpreted as a function, and a function can't be passed as a parameter to the >> operator.
Your problem is the so called most vexing parse. Basically everything, which could be parsed as a function declaration will be parsed as such. Therefore the compiler will interpret int thisisanumber(); as a declaration of a function thisisanumber taking zero arguments and returning an int. If you consider this behaviour the problems with cin>>thisisanumber; should be somewhat selfevident.
If you remove the parantheses, changing the variable declaration to int thisisanumber;, your program should behave like you'd expect it to with thisisanumber being a variable of type int.
You might however reconsider your naming conventions, thisisanumber isn't exactly readable. I would suggest going with this_is_a_number, thisIsANumber or ThisIsANumber.
int thisIsANumber;
try making it variable declaration because what you wrote has been interpreted as function.
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