How do I output a string out if two numbers differs by 1

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How can I output the text when the difference between input1 and input2 is less than 0.1.
Can it includes both negative double and positive double?

C++

#include <iostream>
using namespace std;

int main()
{
double input1=0.00;
    double input2=0.00;
    bool input=true;

    cout<<"Please enter first number"<<std::endl;
    cin>>input1;
    cout<<"Please enter second number"<<std::endl;
    cin>>input2;
    double input_diff=input1-input2;
    cout<<"input difference is "<<input_diff<<endl;

I tried to put this after the code above, but it produces error.

C++

    if (input_diff > -2.0 & < 2.0){      // What is wrong here?
        x=true;
        cout<<"The value between the two is almost equal"<<endl;

    }
}

Posted 25-Aug-15 17:54pm

Are Riff

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Sergey Alexandrovich Kryukov · Accepted Answer · 2015-08-25T18:59:00

Solution 1

Your "& <" is gibberish. Just think about it: is there such operator?
The condition could be (input_diff > -2.0) & (input_diff < 2.0); and if requires any expression on round brackets.

Your questions about 0.1 and negative is just a mess; you write 0.1 and then 2.0, and so on. Please sort it out yourself. Mathematics for elementary school is included in prerequisites for computing, a really smallest part of what you need to know.

—SA

Posted 25-Aug-15 18:59pm

Sergey Alexandrovich Kryukov

Comments

Are Riff 26-Aug-15 5:52am

Your condition works.

if ((input_diff > -2.0) & (input_diff < 2.0)) {
cout << "The value between the two is almost equal" << endl;
}

I never thought of that. Clearly I need more practice.
The reason for 0.1 and 2 is I changed it to see whether double is the reason it fail (and forgot to change it back).

I was reading the if statement like this: if input_diff is larger than -2 and smaller than 2, then cout the string (ignore the x=true;, it was from previous experimenting to make it work).

As for the '&', I copied it from another place, it should be a '&&'. But that didn't work too.

Thank You, I learned something here.

Sergey Alexandrovich Kryukov 26-Aug-15 9:20am

You are welcome.
Good luck, call again.
—SA