Question

Consider the following code:

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

template<class T>
bool IsNaN(T t)
{
    return t != t;
}

int main(int argc, char**argv)
{
    double d1, d2;
    sscanf(argv[1], "%f", &d1);
    sscanf(argv[2], "%f", &d2);

    double dRes = d1/d2;

    cout << "dRes = " << dRes << "\n";

    if(IsNaN(dRes))
        cout << "Is NaN\n";
    else
        cout << "Not NaN\n";

}

Couple of questions:

1. When I pass 0 and 0 as arguments, it outputs dRes = inf. But I was expecting dRes = NaN or something like that.
2. Is NaN representable in double variables? For that matter, any variable?
3. When I changed the data type of d1,d2,dRes to int and passed 0 and 0, I got a Floating exception. What is the difference?
4. How to check if a variable’s value is equal to inf?

Answer 1

1. When using scanf() double should be read using %lf, not %f. %f will convert the input into a 32-bit float, so the first 32 bits of your variables will be filled with some invalid data, and the last 32 bits will be left as garbage.

2. Yes. #include <limits>, then std::numeric_limits<double>::quiet_NaN(). Some compilers (e.g. gcc) also provides the NAN macro in <cmath>.

3. There is no NaN or infinity for integer types. Divide-by-zero for integer will cause an exception (SIGFPE).

4. #include <cmath>, then std::isinf(x). Use std::isfinite(x) to ensure x is not NaN or Infinity.