Why double a=(double)('z'/'A') gives only integer part in C++?

Question

I'm still a beginner in C++ so I seek some help with the basics. Here, in the following code, I'm using type-casting to find value of 122/65 but I'm getting only the integer part even with double data type.

#include <iostream>

using namespace std;

int main()

{

    double a=(double)('z'/'A');

    cout<<a;

    return 0;

}

Can someone provide me a good reason for this??

Thank you.

Because `'z'` and `'A'` are both *integers*, and you have an integer division. You only cast the *result* of the integer division to a `double`. — Some programmer dude, Aug 30 '17 at 07:28
And can you please explain to us why you use *characters* in your division? What is the *actual* problem you attempt to solve by that very odd-looking division? — Some programmer dude, Aug 30 '17 at 07:29

izlin · Accepted Answer · 2017-08-30T07:49:18.923

5

You make an integer division and then you typecast the result to double. Basically you have:

(double) (122/65) = (double) (1) = 1.0
                              ^ truncated -> integer division

If you want a floating point division you can do it this way:

double a = (double)'z' / (double)'A';
//     a = 122.0       / 65.0

edited Aug 30 '17 at 07:49

answered Aug 30 '17 at 07:33

izlin

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@Someprogrammerdude Yes, edited. Thanks. – izlin Aug 30 '17 at 07:49

Why double a=(double)('z'/'A') gives only integer part in C++?

1 Answers1