C++ isinf() function | check if given argument infinite

By Ranjeet V

C++ provides us with a built-in function isinf() that helps us determine whether a given argument is infinite or not. We will see some examples that explain the working of this function further in this tutorial.

C++ isinf() function is defined in <cmath> header. To read more on <cmath> functions, see this: Mathematical functions in C++.

The syntax for isinf() is as follows:

bool isinf( float number);
bool isinf( double number);
bool isinf( long double number);

As you can see, the isinf() function takes a floating-point value as an input parameter. The return type for the function is bool. If the input number is infinite( positive or negative) it returns true i. e. 1 else it returns false i. e. 0.

Let’s understand it with a few examples.

In the given example, we will see what happens when an infinite number is passed as an argument in the function. See the code below.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
  float number = 4.0/0.0;
  if(isinf(number))
  {
    cout << "Number is infinite" << endl;
  }
  else
  {
    cout << "Number is not infinite" << endl;
  }
  
  return 0;
}

The output for the above code is:

Number is infinite

In the above example, we have passed a number (4.0/0.0) as an argument for isinf() function. Since 4.0/0.0 results in infinity, the function returns 1. Therefore, the output print “Number is infinite”.

Now have a look at the below code where we have passed a finite number as input.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
  float number = -4.0;
  if(isinf(number))
  {
    cout << "Number is infinite" << endl;
  }
  else
  {
    cout << "Number is not infinite" << endl;
  }
  
  return 0;
}

Output:

Number is not infinite

As you can see, for a non-infinite number the isinf() function returns 0.

Thank you.

Also read: C++ Math nexttoward() function with example