二进制“ operator！=” |的类型为'double'和'const char [2]'的无效操作数

Question

I have a program that I need to make it so if I enter my first number as x it will terminate but i can't do that because my num1 is a double. If anyone could help me figure out how to do that, that would be great. Thanks

#include <iostream>

using namespace std;
void compute(double, double);

int main()
{
double num1, num2;
while (num1!="x")
{
cout << "Enter First Number: " << endl;
cin >> num1;
cout << "Enter Second Number: " << endl;
cin >> num2;
compute(num1, num2);
}
return 0;
}
void compute(double num1, double num2)
{
double sum,diff,prod,quotient;
cout << "First Number: " << num1 << endl;
cout << "Second Number: " << num2 << endl;
sum=num1+num2;
diff=num1-num2;
prod=num1*num2;
quotient=num1/num2;
cout << "Sum: " << sum << endl;
cout << "Difference: " << diff << endl;
cout << "Product: " << prod << endl;
cout << "Quotient" << quotient << endl;
}

Answer 1

Your main problem here is that you're comparing a number to a const char* (ie, a string literal).

If you want your program to compile, you might want to compare num1 to the character x , for example num1 != 'x' (not single quotes). I will not guarantee this will work correctly, as comparing doubles and characters is a bit ludicrous, but it should compile and may get you a bit further down the road.

Answer 2

You store your input in a std::string and check the content of the string before converting it to a double :

std::string s;
std::cout >> s;

if (s == "x")
  {
     // do something
  }
else
  {
     std::stringstream ss(s);
     double num1;
     ss >> num1;
     // do something else
  }