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isinf() function in C++
  • Last Updated : 27 Sep, 2018

This function is defined in <cmath.h> .The isinf() function is use to determine whether the given number is infinity or not i.e positive infinity or negative infinity both. This function returns 1 if the given number is infinite otherwise this function returns zero.

Syntax:

bool isinf( float arg );

or

bool isinf( double arg );

or

bool isinf( long double arg );

Parameter: This function takes a mandatory parameter x which represents the given floating point value.



Return: This function returns 1 if the given number is infinite else return zero.

Below programs illustrate the isinf() function in C++:

Example 1:- To show infinite case which returns 1

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// c++ program to demonstrate
// example of isnormal() function.
  
#include <bits/stdc++.h>
  
using namespace std;
  
int main()
{
  
    float f = 6.0F;
  
    // check for +ve infinite value
    cout << "isinf(6.0/0.0) is = " << isinf(f/0.0) << endl;
  
    // check for -ve infinite value
    f = -1.2F;
    cout << "isinf(-1.2/0.0) is = " << isinf(f/0.0) << endl;
  
    return 0;
}

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Output:

isinf(6.0/0.0) is = 1
isinf(-1.2/0.0) is = 1

Explanation: In example 1 the floating point number represents infinity that’s why function returns 1.

Example 2:- To show non-infinite case which returns 0

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// c++ program to demonstrate
// example of isinf() function.
  
#include <bits/stdc++.h>
using namespace std;
  
int main()
{
   cout << "isinf(0.0) is = " << isinf(0.0) << endl;
  
   cout << "isinf(sqrt(-1.0)) is = " << isinf(sqrt(-1.0)) << endl;
  
   return 0;
}

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Output:

isinf(0.0) is = 0
isinf(sqrt(-1.0)) is = 0

Exception: In example 2 the given floating point number is not representing infinity that’s why function returns zero.

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